Electricity and Magnetism Practice Questions — JAMB UTME — Page 5

Question 81Question →

Two point charges, $q_1 = +4.0 \times 10^{-8}\text{ C}$ and $q_2 = -9.0 \times 10^{-8}\text{ C}$ , are fixed in a vacuum at a distance of $0.50\text{ m}$ apart. A third point charge $q_3 = +2.0 \times 10^{-8}\text{ C}$ is placed along the line passing through $q_1$ and $q_2$ such that the net electrostatic force acting on it is zero. What is the distance of $q_3$ from $q_1$ ?

0.20\text{ m}

0.40\text{ m}

1.00\text{ m}

1.50\text{ m}

Show answer & explanation

Answer: $1.00\text{ m}$

Answer

The distance of the third charge from

q_1

is

1.00\text{ m}

(located on the side of

q_1

opposite to

q_2

).

The correct distance of

1.00\text{ m}

is determined by recognizing that zero net force on a test charge

q_3

occurs outside the opposite charges

q_1

and

q_2

, specifically on the side of the smaller charge

q_1

. Equating electrostatic forces gives

\frac{4}{d^2} = \frac{9}{(d+0.50)^2}

, which yields

d = 1.00\text{ m}

.

Step-by-Step Solution

1

Determine the equilibrium region for the third charge.

Because

q_1 = +4.0 \times 10^{-8}\text{ C}

and

q_2 = -9.0 \times 10^{-8}\text{ C}

have opposite charges, electrostatic forces on

q_3

point in opposite directions only outside the segment connecting them. To balance the forces,

q_3

must be closer to the smaller magnitude charge

q_1

, placing it at distance

d

to the left of

q_1

.

Between two opposite charges, the force from the positive charge and the force from the negative charge act in the same direction, so net zero force is impossible between them.

2

Set up Coulomb's Law equilibrium equation.

\frac{k |q_1 q_3|}{d^2} = \frac{k |q_2 q_3|}{(d + 0.50)^2} \implies \frac{4.0 \times 10^{-8}}{d^2} = \frac{9.0 \times 10^{-8}}{(d + 0.50)^2}

Equilibrium requires the force magnitude exerted by

q_1

on

q_3

to equal the force magnitude exerted by

q_2

on

q_3

.

3

Solve the equation for distance

d

.

\frac{4}{d^2} = \frac{9}{(d + 0.50)^2} \implies \frac{2}{d} = \frac{3}{d + 0.50} \implies 2(d + 0.50) = 3d \implies d = 1.00\text{ m}

.

Taking the square root of both sides simplifies the inverse-square relation to a solvable linear relation.

Key Concept

Electrostatic equilibrium and vector force cancellation for point charges

Question 82Question →

Two point charges produce mutually perpendicular electric fields at a point $P$ in a vacuum. If the magnitudes of the electric field intensities at $P$ due to the charges individually are $30\text{ N C}^{-1}$ and $40\text{ N C}^{-1}$ , what is the magnitude of the net electric field intensity at point $P$ ?

50\text{ N C}^{-1}

70\text{ N C}^{-1}

10\text{ N C}^{-1}

35\text{ N C}^{-1}

Show answer & explanation

Answer: $50\text{ N C}^{-1}$

Answer

The magnitude of the net electric field intensity at point

P

is

50\text{ N C}^{-1}

.

Because electric field intensity is a vector quantity, two mutually perpendicular electric fields

E_1 = 30\text{ N C}^{-1}

and

E_2 = 40\text{ N C}^{-1}

combine vectorially according to

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = 50\text{ N C}^{-1}

.

Step-by-Step Solution

1

Identify the vector nature and orientation of the given electric fields.

The two component fields

E_1 = 30\text{ N C}^{-1}

and

E_2 = 40\text{ N C}^{-1}

are perpendicular to each other (

\theta = 90^\circ

).

Electric field intensity is a vector quantity, so perpendicular vectors must be combined using vector addition.

2

Apply the Pythagorean theorem to calculate the resultant vector magnitude.

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\text{ N C}^{-1}

.

When two vectors meet at a right angle, the magnitude of their resultant is the hypotenuse of the right-angled triangle formed by the vector components.

Key Concept

Vector Addition of Electric Fields (Superposition Principle)

Estimated Time:1m 0s

Question 83Question →

Two identical point charges, $q_1 = +5.0 \times 10^{-6}\text{ C}$ and $q_2 = +5.0 \times 10^{-6}\text{ C}$ , are fixed in a vacuum at Cartesian coordinates $(0\text{ m}, 3.0\text{ m})$ and $(0\text{ m}, -3.0\text{ m})$ , respectively. A third point charge $q_3 = +2.0 \times 10^{-6}\text{ C}$ is placed on the x-axis at $(4.0\text{ m}, 0\text{ m})$ . Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electrostatic force exerted on $q_3$ ?

5.76 \times 10^{-3}\text{ N}

7.20 \times 10^{-3}\text{ N}

4.32 \times 10^{-3}\text{ N}

3.60 \times 10^{-3}\text{ N}

Show answer & explanation

Answer: $5.76 \times 10^{-3}\text{ N}$

Answer

The magnitude of the net electrostatic force exerted on

q_3

is

5.76 \times 10^{-3}\text{ N}

.

Each charge exerts a repulsive force of magnitude

3.60 \times 10^{-3}\text{ N}

along the line connecting it to the test charge at

(4.0\text{ m}, 0\text{ m})

. Because of symmetry, the y-components of the two forces cancel out completely while their x-components add constructively. Multiplying the individual force magnitude by the cosine of the angle with the x-axis (

\cos\theta = 0.8

) and doubling for both charges gives a net force of

5.76 \times 10^{-3}\text{ N}

.

Step-by-Step Solution

1

Calculate the straight-line distance

r

from

q_1

(or

q_2

) to

q_3

.

r = \sqrt{(4.0 - 0)^2 + (0 - 3.0)^2} = \sqrt{16 + 9} = 5.0\text{ m}

.

Coulomb's Law requires the straight-line separation distance between interacting point charges.

2

Calculate the magnitude of the electrostatic force

F_1

exerted on

q_3

by

q_1

.

F_1 = \frac{k \cdot q_1 \cdot q_3}{r^2} = \frac{(9.0 \times 10^9) \cdot (5.0 \times 10^{-6}) \cdot (2.0 \times 10^{-6})}{5.0^2} = \frac{9.0 \times 10^{-2}}{25} = 3.60 \times 10^{-3}\text{ N}

.

By symmetry, the force magnitude

F_2

from

q_2

on

q_3

is also equal to

3.60 \times 10^{-3}\text{ N}

.

3

Determine the vector components of the forces along the axes.

The cosine of the angle

\theta

with the positive x-axis is

\cos\theta = \frac{4.0}{5.0} = 0.8

. The sine is

\sin\theta = \frac{3.0}{5.0} = 0.6

. The vertical y-components are equal in magnitude and opposite in direction (

F_{1y} = -F_{2y}

), canceling to zero.

Forces are vector quantities; symmetrically placed identical charges produce opposing vertical components and reinforcing horizontal components.

4

Sum the horizontal x-components to obtain the net force.

F_{\text{net}} = F_{1x} + F_{2x} = 2 \cdot F_1 \cos\theta = 2 \cdot (3.60 \times 10^{-3}\text{ N}) \cdot 0.8 = 5.76 \times 10^{-3}\text{ N}

.

Both x-components point along the positive x-axis, so their magnitudes add directly.

Key Concept

Vector superposition of electric forces

Question 84Question →

Three point charges $q_1 = +2.0 \times 10^{-6}\text{ C}$ , $q_2 = +2.0 \times 10^{-6}\text{ C}$ , and $q_3 = -4.0 \times 10^{-6}\text{ C}$ are placed along a straight line at positions $x = 0\text{ m}$ , $x = 0.30\text{ m}$ , and $x = 0.60\text{ m}$ , respectively. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , calculate the magnitude of the net electrostatic force acting on charge $q_2$ in newtons.

Show answer & explanation

Answer: 1.2

Answer

The magnitude of the net electrostatic force acting on charge

q_2

is

1.20\text{ N}

.

The force exerted on

q_2

by

q_1

is repulsive (

0.40\text{ N}

directed to the right) because both charges are positive. The force exerted on

q_2

by

q_3

is attractive (

0.80\text{ N}

directed to the right) because

q_2

is positive and

q_3

is negative. Since both component forces act in the same direction, the total net force magnitude is

0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

.

Step-by-Step Solution

1

Calculate the repulsive force exerted by

q_1

on

q_2

F_{12} = 0.40\text{ N}

pointing to the right

Like charges repel each other, so

q_1

pushes

q_2

away along the

+x

-axis.

2

Calculate the attractive force exerted by

q_3

on

q_2

F_{32} = 0.80\text{ N}

pointing to the right

Unlike charges attract each other, so

q_3

pulls

q_2

towards itself along the

+x

-axis.

3

Sum the component electrostatic forces acting on

q_2

F_{\text{net}} = 0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Because both forces act in the exact same direction along the line, their magnitudes add directly.

Key Concept

Coulomb's Law and Principle of Superposition for Electrostatic Forces

Question 85Question →

Two capacitors with capacitances of $10\text{ }\mu\text{F}$ and $15\text{ }\mu\text{F}$ are connected in parallel. What is the equivalent capacitance of the combination in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 25

Answer

The equivalent capacitance of the parallel combination is

25\text{ }\mu\text{F}

.

When capacitors are connected in parallel, the total equivalent capacitance is equal to the direct sum of the individual capacitances:

C_{\text{eq}} = C_1 + C_2 = 10\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 25\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the relationship for parallel capacitors

C_{\text{eq}} = C_1 + C_2

Capacitors connected in parallel store charge independently across the same potential difference, so their capacitances add directly.

2

Substitute the given values into the formula

C_{\text{eq}} = 10\text{ }\mu\text{F} + 15\text{ }\mu\text{F}

The circuit contains two capacitors of

10\text{ }\mu\text{F}

and

15\text{ }\mu\text{F}

in parallel.

3

Calculate the total capacitance

25\text{ }\mu\text{F}

Simple addition of the two values yields

25\text{ }\mu\text{F}

.

Key Concept

Equivalent Capacitance of Parallel Connected Capacitors

Question 86Question →

What is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a point $2.0\text{ m}$ away from a isolated point charge of $+4.0 \times 10^{-6}\text{ C}$ in a vacuum? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

Show answer & explanation

Answer: 9000

Answer

The magnitude of the electric field intensity is

9000\text{ N C}^{-1}

.

The electric field intensity

E

produced by a point charge

q

at distance

r

is given by

E = \frac{kq}{r^2}

. Substituting

k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}

,

q = 4.0 \times 10^{-6}\text{ C}

, and

r = 2.0\text{ m}

yields

E = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-6}}{4.0} = 9000\text{ N C}^{-1}

.

Step-by-Step Solution

1

Identify the given physical quantities and formula

q = 4.0 \times 10^{-6}\text{ C}

,

r = 2.0\text{ m}

,

k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}

. Formula:

E = \frac{kq}{r^2}

The magnitude of electric field intensity due to a single point charge is given by Coulomb's field law.

2

Substitute the values and calculate the electric field strength

E = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-6}}{2.0^2} = \frac{36000}{4} = 9000\text{ N C}^{-1}

Perform basic arithmetic simplification to determine the numerical result.

Key Concept

Electric Field Intensity due to a Point Charge

Question 87Question →

Two identical isolated metal spheres, $X$ and $Y$ , carry initial charges of $+q$ and $-3q$ respectively and are separated by a fixed distance $r$ in a vacuum. The magnitude of the electrostatic force between them is $F$ . A third identical, uncharged metal sphere $Z$ is touched briefly to sphere $X$ , then touched briefly to sphere $Y$ , and finally placed at the midpoint between spheres $X$ and $Y$ . What is the magnitude of the net electrostatic force acting on sphere $Z$ in terms of $F$ ?

\dfrac{35}{12}F

\dfrac{5}{4}F

\dfrac{35}{48}F

\dfrac{35}{4}F

Show answer & explanation

Answer: $\dfrac{35}{12}F$

Answer

The magnitude of the net electrostatic force acting on sphere

Z

is

\dfrac{35}{12}F

.

When sphere Z touches sphere X, charge is shared equally so both carry

+\frac{q}{2}

. Next, when sphere Z touches sphere Y (charge

-3q

), total charge becomes

-\frac{5q}{2}

, dividing equally into

-\frac{5q}{4}

for each. At the midpoint (

r/2

from each sphere), sphere X (

+\frac{q}{2}

) attracts sphere Z (

-\frac{5q}{4}

) towards the left with force

\frac{5}{2}\frac{kq^2}{r^2}

. Sphere Y (

-\frac{5q}{4}

) repels sphere Z (

-\frac{5q}{4}

) towards the left with force

\frac{25}{4}\frac{kq^2}{r^2}

. Adding these co-directional forces yields

\frac{35}{4}\frac{kq^2}{r^2}

. Given the initial force

F = \frac{3kq^2}{r^2}

, we substitute

\frac{kq^2}{r^2} = \frac{F}{3}

to obtain

\frac{35}{12}F

.

Step-by-Step Solution

1

Determine the initial electrostatic force

F

between spheres

X

and

Y

.

F = k \frac{|(+q)(-3q)|}{r^2} = \frac{3kq^2}{r^2}

, which gives

\frac{kq^2}{r^2} = \frac{F}{3}

.

Coulomb's law defines force as proportional to the product of charges divided by the square of separation distance.

2

Calculate the charges on the spheres after sequential contacts.

When

Z

(

0

) touches

X

(

+q

), charge divides equally:

q_X' = +\frac{q}{2}

and

q_Z' = +\frac{q}{2}

. When

Z

(

+\frac{q}{2}

) touches

Y

(

-3q

), the combined charge is

+\frac{q}{2} - 3q = -\frac{5q}{2}

, which divides equally to give

q_Y' = -\frac{5q}{4}

and

q_Z'' = -\frac{5q}{4}

.

Identical conductors share total charge equally upon contact due to conservation of charge and symmetric potential.

3

Determine the forces exerted on sphere

Z

at the midpoint.

Separation distance from

Z

to both

X

and

Y

is

\frac{r}{2}

. Force from

X

on

Z

(attractive, pulling towards

X

):

F_{ZX} = k \frac{|(+\frac{q}{2})(-\frac{5q}{4})|}{(\frac{r}{2})^2} = k \frac{\frac{5q^2}{8}}{\frac{r^2}{4}} = \frac{5}{2}\frac{kq^2}{r^2}

. Force from

Y

on

Z

(repulsive, pushing away from

Y

toward

X

):

F_{ZY} = k \frac{|(-\frac{5q}{4})(-\frac{5q}{4})|}{(\frac{r}{2})^2} = k \frac{\frac{25q^2}{16}}{\frac{r^2}{4}} = \frac{25}{4}\frac{kq^2}{r^2}

.

Opposite charges attract and like charges repel. Midpoint separation distance is

r/2

.

4

Calculate the net force on

Z

and express it in terms of

F

.

Since both forces act in the same direction (towards sphere

X

),

F_{\text{net}} = F_{ZX} + F_{ZY} = (\frac{5}{2} + \frac{25}{4})\frac{kq^2}{r^2} = \frac{35}{4}\frac{kq^2}{r^2}

. Substituting

\frac{kq^2}{r^2} = \frac{F}{3}

yields

F_{\text{net}} = \frac{35}{4} \times \frac{F}{3} = \frac{35}{12}F

.

Forces in the same direction add vectorially.

Key Concept

Electrostatic Charge Sharing and Coulomb's Law Vector Superposition

Question 88Question →

Two capacitors of capacitances $C_1 = 6\text{ }\mu\text{F}$ and $C_2 = 12\text{ }\mu\text{F}$ are connected in series across a $180\text{ V}$ direct current power supply. After the capacitors are fully charged, the supply is disconnected. A dielectric material of dielectric constant $K = 4$ is then inserted to completely fill the space between the plates of $C_1$ . What is the new potential difference across $C_1$ ?

30\text{ V}

60\text{ V}

135\text{ V}

15\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

30\text{ V}

The initial equivalent capacitance of the series combination is

4\text{ }\mu\text{F}

, which charges each capacitor to

720\text{ }\mu\text{C}

. Because the circuit is disconnected from the battery, charge is conserved. Inserting a dielectric of constant

K = 4

increases

C_1

to

24\text{ }\mu\text{F}

, resulting in a potential difference of

V = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the initial equivalent capacitance of the series network

C_{eq} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

.

2

Determine the charge stored on each capacitor prior to disconnection

Q = 720\text{ }\mu\text{C}

Total charge supplied by the

180\text{ V}

battery is

Q = C_{eq} V = 4\text{ }\mu\text{F} \times 180\text{ V} = 720\text{ }\mu\text{C}

. In series, each capacitor holds this same charge.

3

Calculate the modified capacitance of

C_1

after dielectric insertion

C_1' = 24\text{ }\mu\text{F}

A dielectric of constant

K = 4

scales capacitance by

K

:

C_1' = K \times C_1 = 4 \times 6\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

.

4

Calculate the final potential difference across

C_1

using charge conservation

V_1' = 30\text{ V}

Because the source is disconnected, charge

Q = 720\text{ }\mu\text{C}

on

C_1

remains constant. Therefore,

V_1' = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Key Concept

Effect of dielectrics and charge conservation in disconnected series capacitor circuits

Question 89Question →

Two identical positive point charges, each of magnitude $q = +2.5 \times 10^{-6}\text{ C}$ , are fixed in a vacuum at a distance of $0.60\text{ m}$ apart. A third point charge $q_0 = +1.0 \times 10^{-6}\text{ C}$ is placed on the perpendicular bisector of the line joining the two fixed charges, at a distance of $0.40\text{ m}$ from their midpoint. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electrostatic force acting on the third charge, in newtons?

Show answer & explanation

Answer: 0.144

Answer

The magnitude of the net electrostatic force acting on the third charge is

0.144\text{ N}

.

Each fixed charge exerts an equal repulsive electrostatic force of

0.09\text{ N}

on the third charge. Due to the symmetrical arrangement, the force components perpendicular to the bisector cancel each other out, while the parallel components add together, giving a net force of

2 \times 0.09 \times 0.8 = 0.144\text{ N}

.

Step-by-Step Solution

1

Determine the distance from each fixed charge to the third charge

r = 0.50\text{ m}

The charges form a right-angled triangle with base

0.30\text{ m}

(half of

0.60\text{ m}

) and height

0.40\text{ m}

, yielding a hypotenuse of

\sqrt{0.30^2 + 0.40^2} = 0.50\text{ m}

.

2

Calculate the magnitude of the individual repulsive force from one charge

F = 0.09\text{ N}

Applying Coulomb's law:

F = \frac{k q q_0}{r^2} = \frac{9.0 \times 10^9 \times 2.5 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.25} = 0.09\text{ N}

.

3

Determine directional component of forces along the perpendicular bisector

\cos\theta = 0.8

The directional cosine along the axis of symmetry is the ratio of the adjacent side (

0.40\text{ m}

) to the hypotenuse (

0.50\text{ m}

).

4

Compute the net electrostatic force using vector addition

F_{\text{net}} = 0.144\text{ N}

Horizontal components cancel by symmetry, so

F_{\text{net}} = 2 F \cos\theta = 2 \times 0.09 \times 0.8 = 0.144\text{ N}

.

Key Concept

Vector superposition of Coulombic forces along an axis of symmetry

Question 90Question →

A parallel plate air capacitor of capacitance $8.0\text{ }\mu\text{F}$ is fully charged using a $40.0\text{ V}$ d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of $4.0$ is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules ( $\mu\text{J}$ ).

Show answer & explanation

Answer: 4800

Answer

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.

Disconnecting the battery ensures that the charge

Q = C_0 V = 320\text{ }\mu\text{C}

on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to

32.0\text{ }\mu\text{F}

. The energy decreases from

U_i = 6400\text{ }\mu\text{J}

to

U_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}

, giving a total decrease of

4800\text{ }\mu\text{J}

.

Step-by-Step Solution

1

Calculate the initial energy stored in the air capacitor before disconnection.

Initial energy

U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}

.

The initial state has known capacitance and potential difference.

2

Calculate the new capacitance with the dielectric present.

Final capacitance

C_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}

.

Inserting a dielectric of constant

K

scales the capacitance by

K

.

3

Calculate the final stored energy using charge conservation.

Final energy

U_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}

.

Disconnection forces the charge

Q

to remain fixed, so energy scales inversely with capacitance (

U = \frac{Q^2}{2C}

).

4

Find the difference between initial and final energy.

\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.

The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Key Concept

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions

Estimated Time:2m 0s

Question 91Question →

An electric charge of $+3.2 \times 10^{-19}\text{ C}$ is situated in a uniform electric field of strength $5.0 \times 10^4\text{ N C}^{-1}$ . What is the magnitude of the electrostatic force exerted on the charge?

6.4 \times 10^{-24}\text{ N}

1.6 \times 10^{-14}\text{ N}

1.6 \times 10^{-23}\text{ N}

5.0 \times 10^4\text{ N}

Show answer & explanation

Answer: $1.6 \times 10^{-14}\text{ N}$

Answer

The magnitude of the electrostatic force exerted on the charge is

1.6 \times 10^{-14}\text{ N}

.

The relationship between electric field intensity

E

, charge

q

, and electrostatic force

F

is given by

F = qE

. Multiplying

+3.2 \times 10^{-19}\text{ C}

by

5.0 \times 10^4\text{ N C}^{-1}

gives

1.6 \times 10^{-14}\text{ N}

, which correctly expresses the force magnitude in standard notation.

Step-by-Step Solution

1

Identify the given values and formula.

Charge

q = 3.2 \times 10^{-19}\text{ C}

, Electric field strength

E = 5.0 \times 10^4\text{ N C}^{-1}

. Formula:

F = qE

.

The force experienced by a charge in an electric field is the product of the charge magnitude and the field strength.

2

Substitute the values into the formula and calculate.

F = (3.2 \times 10^{-19}) \times (5.0 \times 10^4) = 16.0 \times 10^{-15}\text{ N} = 1.6 \times 10^{-14}\text{ N}

.

Multiplying the numerical coefficients

(3.2 \times 5.0 = 16.0)

and combining the powers of ten

(10^{-19} \times 10^4 = 10^{-15})

gives

1.6 \times 10^{-14}\text{ N}

in standard scientific notation.

Key Concept

Electric Field Intensity and Electrostatic Force (

F = qE

)

Question 92Question →

A parallel-plate capacitor with air between its plates has a capacitance of $12\text{ }\mu\text{F}$ . A dielectric slab of relative permittivity $\varepsilon_r = 4$ and thickness $t = \frac{d}{3}$ , where $d$ is the total plate separation, is inserted between the plates parallel to them. What is the effective capacitance of the modified capacitor?

16\text{ }\mu\text{F}

48\text{ }\mu\text{F}

162\text{ }\mu\text{F}

6\text{ }\mu\text{F}

Show answer & explanation

Answer: $16\text{ }\mu\text{F}$

Answer

The effective capacitance of the modified capacitor is

16\text{ }\mu\text{F}

.

Inserting a dielectric slab of thickness

t = d/3

creates a system equivalent to two series capacitors: an air-filled region of thickness

2d/3

(

C_1 = 1.5 C_0 = 18\text{ }\mu\text{F}

) and a dielectric-filled region of thickness

d/3

(

C_2 = 12 C_0 = 144\text{ }\mu\text{F}

). Combining them via the series reciprocal formula gives

C_{\text{eq}} = \frac{18 \times 144}{18 + 144} = 16\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Model the partially filled capacitor as two capacitors connected in series.

Air layer of thickness

d_1 = d - t = \frac{2}{3}d

forms capacitor

C_1

. Dielectric layer of thickness

d_2 = t = \frac{1}{3}d

forms capacitor

C_2

.

Dividing the plate gap vertically into two distinct media creates two capacitive regions sharing the same electric flux path.

2

Calculate the individual capacitances

C_1

and

C_2

in terms of initial air capacitance

C_0 = 12\text{ }\mu\text{F}

.

C_1 = \frac{\varepsilon_0 A}{\frac{2}{3}d} = \frac{3}{2}C_0 = \frac{3}{2}(12) = 18\text{ }\mu\text{F}

and

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{\frac{1}{3}d} = 3 \varepsilon_r C_0 = 3(4)(12) = 144\text{ }\mu\text{F}

.

Capacitance is inversely proportional to plate distance and directly proportional to relative permittivity.

3

Calculate the equivalent capacitance

C_{\text{eq}}

for two series capacitors.

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{18} + \frac{1}{144} = \frac{8 + 1}{144} = \frac{9}{144} = \frac{1}{16}\text{ }\mu\text{F}^{-1} \implies C_{\text{eq}} = 16\text{ }\mu\text{F}

.

Capacitors connected in series combine reciprocally.

Key Concept

Partially filled parallel-plate capacitors act as series combinations of distinct capacitive layers.

Estimated Time:3m 0s

Question 93Question →

Two capacitors with capacitances of $3.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel across a direct-current source. What is the equivalent capacitance of this combination?

9.0\text{ }\mu\text{F}

2.0\text{ }\mu\text{F}

4.5\text{ }\mu\text{F}

18.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $9.0\text{ }\mu\text{F}$

Answer

The equivalent capacitance of the parallel combination is

9.0\text{ }\mu\text{F}

.

When capacitors are connected in parallel, each capacitor experiences the full potential difference of the voltage source, and the total charge stored is the sum of individual charges (

Q_{total} = Q_1 + Q_2

). Thus, the equivalent capacitance is the direct sum of the individual capacitances:

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the combination rule for parallel capacitors.

The total capacitance is given by

C_{eq} = C_1 + C_2

.

Capacitors in parallel share the same potential difference, so total charge stored is the sum of individual charges.

2

Substitute the given values into the parallel capacitance formula.

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Direct addition yields the total equivalent capacitance.

Key Concept

Parallel Combination of Capacitors

Estimated Time:45s

Question 94Question →

Two identical air-filled parallel-plate capacitors, $C_1$ and $C_2$ , each of capacitance $C$ , are connected in series across a direct-current voltage source of potential difference $V$ . While the circuit remains connected to the voltage source, a dielectric slab of relative permittivity $\varepsilon_r = 3$ is fully inserted into $C_1$ , completely filling the space between its plates. What is the ratio of the electrostatic energy stored in $C_1$ after inserting the dielectric to the energy stored in $C_1$ before the insertion?

3 : 4

1 : 3

3 : 1

27 : 4

Show answer & explanation

Answer: $3 : 4$

Answer

The ratio of the energy stored in the first capacitor after dielectric insertion to before insertion is

3 : 4

(or

0.75

).

Initially, the two identical capacitors divide the total source voltage

V

equally, giving

V_1 = V/2

and initial energy

U_{1,i} = \frac{1}{8} C V^2

. When the dielectric of relative permittivity

3

is inserted, the capacitance of the first capacitor becomes

3C

. In a series circuit connected to a constant voltage source, the total charge becomes

Q = C_{eq}V = \frac{3}{4}CV

, which reduces the voltage across the modified capacitor to

V_1' = \frac{Q}{3C} = \frac{V}{4}

. The new stored energy is

U_{1,f} = \frac{1}{2}(3C)(\frac{V}{4})^2 = \frac{3}{32} C V^2

. Dividing

U_{1,f}

by

U_{1,i}

yields

\frac{3/32}{1/8} = \frac{3}{4}

.

Step-by-Step Solution

1

Calculate initial capacitance and voltage across

C_1

Initial capacitance

C_{1,i} = C

. Since

C_1

and

C_2

are identical and in series, initial potential difference across

C_1

is

V_{1,i} = \frac{V}{2}

.

Equal capacitors in series divide total voltage equally.

2

Calculate initial electrostatic energy stored in

C_1

U_{1,i} = \frac{1}{2} C_{1,i} V_{1,i}^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{8} C V^2

.

Formula for energy stored in a capacitor is

U = \frac{1}{2} C V^2

.

3

Determine final capacitance of

C_1

and new voltage division

New capacitance

C_{1,f} = \varepsilon_r C = 3C

. Total equivalent capacitance

C_{eq} = \frac{3C \cdot C}{3C + C} = \frac{3}{4} C

. Total charge supplied

Q = C_{eq} V = \frac{3}{4} C V

. Final voltage across

C_1

is

V_{1,f} = \frac{Q}{C_{1,f}} = \frac{\frac{3}{4} C V}{3C} = \frac{V}{4}

.

Dielectric increases capacitance by factor

\varepsilon_r

, altering equivalent capacitance and potential distribution in series.

4

Calculate final electrostatic energy in

C_1

and compute the ratio

U_{1,f} = \frac{1}{2} (3C) \left(\frac{V}{4}\right)^2 = \frac{3}{32} C V^2

. Ratio

\frac{U_{1,f}}{U_{1,i}} = \frac{\frac{3}{32} C V^2}{\frac{1}{8} C V^2} = \frac{3}{4}

.

Divide final stored energy by initial stored energy.

Key Concept

Series combination of capacitors with dielectric insertion under constant battery voltage

Question 95Question →

Three capacitors, each of capacitance $12\text{ }\mu\text{F}$ , are arranged such that two of them are connected in parallel, and this combination is connected in series with the third capacitor. If the entire network is connected across a $20\text{ V}$ d.c. power supply, what is the total energy stored in the network?

0.8 \times 10^{-3}\text{ J}

1.6 \times 10^{-3}\text{ J}

3.6 \times 10^{-3}\text{ J}

7.2 \times 10^{-3}\text{ J}

Show answer & explanation

Answer: $1.6 \times 10^{-3}\text{ J}$

Answer

The total energy stored in the network is

1.6 \times 10^{-3}\text{ J}

.

Combining two

12\text{ }\mu\text{F}

capacitors in parallel gives a parallel capacitance of

24\text{ }\mu\text{F}

. Connecting this combination in series with the third

12\text{ }\mu\text{F}

capacitor results in an equivalent network capacitance of

C_{\text{eq}} = \frac{24 \times 12}{24 + 12} = 8\text{ }\mu\text{F}

. Substituting this into the energy formula

E = \frac{1}{2} C_{\text{eq}} V^2

yields

E = \frac{1}{2} \times 8 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel branch.

C_p = 12\text{ }\mu\text{F} + 12\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network.

C_{\text{eq}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{24 \times 12}{24 + 12} = \frac{288}{36} = 8\text{ }\mu\text{F} = 8 \times 10^{-6}\text{ F}

The parallel combination is connected in series with the third capacitor.

3

Calculate the total electrical energy stored in the combination.

E = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times (8 \times 10^{-6}\text{ F}) \times (20\text{ V})^2 = 4 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

The formula for energy stored in a capacitor network is

E = \frac{1}{2} C V^2

.

Key Concept

Mixed capacitor networks and energy storage

Estimated Time:1m 30s

Question 96Question →

Two electrostatic forces of magnitude $3.0\text{ N}$ and $4.0\text{ N}$ act on a small test charge at right angles ( $90^\circ$ ) to each other. What is the magnitude of the net electrostatic force acting on the charge?

7.0\text{ N}

5.0\text{ N}

1.0\text{ N}

12.0\text{ N}

Show answer & explanation

Answer: $5.0\text{ N}$

Answer

5.0\text{ N}

The net electrostatic force is found by vector addition. Since the two forces are perpendicular (

90^\circ

), the magnitude of their resultant is given by

\sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = 5.0\text{ N}

.

Step-by-Step Solution

1

Identify the nature of force as a vector quantity.

Electrostatic forces must be combined using vector addition rather than simple scalar addition.

The forces act at right angles (

90^\circ

) to each other.

2

Apply the Pythagorean theorem to calculate the magnitude of the resultant net force

F_{\text{net}}

.

F_{\text{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\text{ N}

.

For perpendicular vectors, the resultant is the hypotenuse of a right-angled triangle formed by the vector components.

Key Concept

Vector Addition of Electrostatic Forces

Question 97Question →

Two point charges, $q_1 = +2.0 \times 10^{-6}\text{ C}$ and $q_2 = -2.0 \times 10^{-6}\text{ C}$ , are fixed in a vacuum separated by a distance of $0.20\text{ m}$ . What is the magnitude of the net electric field intensity at the midpoint along the line joining the two charges? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

0.0\text{ N C}^{-1}

9.0 \times 10^5\text{ N C}^{-1}

1.8 \times 10^6\text{ N C}^{-1}

3.6 \times 10^6\text{ N C}^{-1}

Show answer & explanation

Answer: $3.6 \times 10^6\text{ N C}^{-1}$

Answer

3.6 \times 10^6\text{ N C}^{-1}

The correct answer is

3.6 \times 10^6\text{ N C}^{-1}

. At the midpoint (

r = 0.10\text{ m}

), the electric field due to the positive charge points towards the negative charge, and the field due to the negative charge also points towards the negative charge. Summing both equal field magnitudes of

1.8 \times 10^6\text{ N C}^{-1}

yields

3.6 \times 10^6\text{ N C}^{-1}

.

Step-by-Step Solution

1

Determine the distance from each point charge to the midpoint.

The distance

r = \frac{0.20\text{ m}}{2} = 0.10\text{ m}

.

The midpoint divides the total separation distance equally.

2

Calculate the magnitude of the electric field intensity

E_1

created by the positive charge

q_1

at the midpoint.

E_1 = \frac{k |q_1|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1}

directed away from

q_1

(towards

q_2

).

Electric field vectors point away from positive charges.

3

Calculate the magnitude of the electric field intensity

E_2

created by the negative charge

q_2

at the midpoint.

E_2 = \frac{k |q_2|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1}

directed towards

q_2

.

Electric field vectors point towards negative charges.

4

Combine the electric field vectors vectorially to find the net field intensity.

E_{\text{net}} = E_1 + E_2 = 1.8 \times 10^6 + 1.8 \times 10^6 = 3.6 \times 10^6\text{ N C}^{-1}

.

Because both

E_1

and

E_2

point in the exact same direction (towards

q_2

), their magnitudes add directly.

Key Concept

Superposition Principle of Electric Fields

Question 98Question →

In an electric circuit, two capacitors $C_1 = 12\text{ }\mu\text{F}$ and $C_2 = 6\text{ }\mu\text{F}$ are connected in series. This series combination is then connected in parallel with a third capacitor $C_3$ of unknown value. When a direct-current potential difference of $100\text{ V}$ is applied across the entire network, the total electrostatic energy stored in the circuit is $100\text{ mJ}$ . What is the capacitance of $C_3$ in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 16

Answer

The capacitance of

C_3

is

16\text{ }\mu\text{F}

.

First, the series combination of

12\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

yields an equivalent branch capacitance of

4\text{ }\mu\text{F}

. Second, using

E = \frac{1}{2} C_{eq} V^2

with

E = 0.100\text{ J}

and

V = 100\text{ V}

gives a total circuit equivalent capacitance of

20\text{ }\mu\text{F}

. Finally, subtracting the branch capacitance from the total parallel equivalent capacitance gives

C_3 = 20\text{ }\mu\text{F} - 4\text{ }\mu\text{F} = 16\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the effective capacitance of the series branch containing

C_1

and

C_2

C_{12} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} \implies C_{12} = 4\text{ }\mu\text{F}

.

2

Determine the total equivalent capacitance

C_{eq}

of the circuit using the given stored energy and voltage

C_{eq} = 20\text{ }\mu\text{F}

Energy stored in a capacitor network is

E = \frac{1}{2} C_{eq} V^2

. Rearranging gives

C_{eq} = \frac{2E}{V^2} = \frac{2 \times 0.100\text{ J}}{(100\text{ V})^2} = 20 \times 10^{-6}\text{ F} = 20\text{ }\mu\text{F}

.

3

Calculate the unknown capacitance

C_3

from the parallel combination formula

C_3 = 16\text{ }\mu\text{F}

Because the branch

C_{12}

and

C_3

are in parallel,

C_{eq} = C_{12} + C_3 \implies 20\text{ }\mu\text{F} = 4\text{ }\mu\text{F} + C_3 \implies C_3 = 16\text{ }\mu\text{F}

.

Key Concept

Series and parallel combinations of capacitors combined with electrostatic energy storage

Estimated Time:2m 0s

Question 99Question →

A neutral insulated conductor loses $5.0 \times 10^{13}$ electrons during an electrostatics experiment. What is the magnitude of the net electric charge, in microcoulombs ( $\mu\text{C}$ ), acquired by the conductor? (Take elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ )

Show answer & explanation

Answer: 8

Answer

The magnitude of the net electric charge acquired by the conductor is

8\text{ }\mu\text{C}

.

According to the principle of charge quantization, the total electric charge

Q

is calculated using

Q = n e

. Multiplying

5.0 \times 10^{13}

electrons by

1.6 \times 10^{-19}\text{ C}

yields

8.0 \times 10^{-6}\text{ C}

, which converts to

8\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Identify the relevant formula for quantization of charge.

The net charge acquired is given by

Q = n e

.

Electric charge is quantized, so the total charge magnitude equals the number of transferred electrons multiplied by the magnitude of charge on a single electron.

2

Calculate the magnitude of charge in Coulombs.

Q = (5.0 \times 10^{13}) \times (1.6 \times 10^{-19}\text{ C}) = 8.0 \times 10^{-6}\text{ C}

.

Multiplying the quantity of removed electrons by the elementary charge gives total charge in Coulombs.

3

Convert the calculated value from Coulombs to microcoulombs.

8.0 \times 10^{-6}\text{ C} = 8\text{ }\mu\text{C}

.

Since

1\text{ }\mu\text{C} = 10^{-6}\text{ C}

, dividing

8.0 \times 10^{-6}

by

10^{-6}

yields 8.

Key Concept

Quantization of Electric Charge

Question 100Question →

A $5.0\text{ }\mu\text{F}$ parallel-plate capacitor is connected across a potential difference of $20.0\text{ V}$ . What is the magnitude of the electric charge stored on either plate of the capacitor, in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 100

Answer

The magnitude of the electric charge stored on either plate of the capacitor is

100.0\text{ }\mu\text{C}

.

Using the capacitor charge equation

Q = C \times V

, substituting

C = 5.0\text{ }\mu\text{F}

and

V = 20.0\text{ V}

yields

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Identify the given physical quantities and the formula for electric charge on a capacitor.

Capacitance

C = 5.0\text{ }\mu\text{F}

, voltage

V = 20.0\text{ V}

. Formula:

Q = C V

.

The charge stored by a capacitor is directly proportional to the potential difference across its terminals.

2

Perform the multiplication to determine the charge magnitude in microcoulombs.

Q = 5.0 \times 20.0 = 100.0\text{ }\mu\text{C}

.

Multiplying capacitance in microfarads by potential difference in volts yields charge directly in microcoulombs.

Key Concept

Fundamental relationship between capacitance, charge, and potential difference (

Q = C V

)

Estimated Time:45s

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