Electrostatics and Electric Charges Practice Questions — JAMB UTME

Question 1Question →

Two point charges of $+2.0 \times 10^{-6}\text{ C}$ and $+4.0 \times 10^{-6}\text{ C}$ are placed in a vacuum at a distance of $0.3\text{ m}$ apart. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force exerted between them in Newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the electrostatic force between the charges is 0.8 N.

According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them:

F = \frac{k q_1 q_2}{r^2}

. Substituting

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

, and

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

into the formula gives

F = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify known quantities from the problem statement

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

,

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

Clear identification of parameters is required for substitution into Coulomb's Law.

2

Apply Coulomb's Law formula

F = \frac{k q_1 q_2}{r^2}

Coulomb's Law quantifies the electrostatic force between two stationary point charges.

3

Substitute values and perform arithmetic calculation

F = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}

Squaring the separation distance

0.3\text{ m}

gives

0.09\text{ m}^2

, and evaluating the numerator gives

0.072\text{ N}\cdot\text{m}^2

.

Key Concept

Coulomb's Law

Question 2Question →

Two equal positive point charges, each of magnitude $2.0 \times 10^{-6}\text{ C}$ , are placed $0.2\text{ m}$ apart in a vacuum. What is the magnitude of the net electric field intensity at the midpoint between the two charges? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

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

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

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

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

Show answer & explanation

Answer: $0\text{ N C}^{-1}$

Answer

The net electric field intensity at the midpoint is

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

.

At the midpoint between two identical positive charges, the electric field created by each charge has the exact same magnitude because the charges and distances are equal. Because electric field lines point away from positive charges, the two field vectors at the midpoint point in directly opposite directions. Taking vector superposition gives a net electric field of zero.

Step-by-Step Solution

1

Determine the distance from each charge to the midpoint.

The midpoint distance

r = \frac{0.2\text{ m}}{2} = 0.1\text{ m}

.

Electric field calculation requires the distance from the point charge to the point of evaluation.

2

Calculate the magnitude of the electric field due to one charge.

E = \frac{k Q}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.1)^2} = 1.8 \times 10^6\text{ N C}^{-1}

.

Electric field intensity magnitude is given by Coulomb's field formula

E = \frac{k Q}{r^2}

.

3

Apply vector addition to find the net electric field at the midpoint.

E_{\text{net}} = E_1 - E_2 = 1.8 \times 10^6 - 1.8 \times 10^6 = 0\text{ N C}^{-1}

.

Electric field is a vector quantity. Since both charges are positive, the field vectors point away from each charge and act in opposite directions at the midpoint.

Key Concept

Vector superposition of electric fields

Question 3Question →

A point charge $q_1 = +9.0 \times 10^{-9}\text{ C}$ is fixed at the origin ( $x = 0\text{ m}$ ), and a second point charge $q_2 = -4.0 \times 10^{-9}\text{ C}$ is fixed on the x-axis at $x = 0.5\text{ m}$ . At what position $x$ (in meters) along the x-axis is the net electric field intensity equal to zero?

Show answer & explanation

Answer: 1.5

Answer

The net electric field intensity is zero at

x = 1.5\text{ m}

.

The correct position is

x = 1.5\text{ m}

. At this point, the electric field from

+q_1

points in the

+x

direction with magnitude

E_1 = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-9}}{1.5^2} = 36\text{ N C}^{-1}

, and the electric field from

-q_2

points in the

-x

direction with magnitude

E_2 = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{1.0^2} = 36\text{ N C}^{-1}

. The two vectors are equal in magnitude and opposite in direction, yielding a net electric field of zero.

Step-by-Step Solution

1

Determine the physical region where electric fields can cancel

The point of zero field lies to the right of

q_2

, i.e.,

x > 0.5\text{ m}

.

Between the charges, the fields due to

+q_1

and

-q_2

point in the same direction (+x). To the left of

q_1

,

q_1

is both larger in magnitude and closer, so

E_1 > E_2

everywhere. Hence, balance can only occur to the right of the smaller magnitude charge

q_2

.

2

Set up the condition for equal electric field magnitudes

\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(x - 0.5)^2}

For the net field to be zero, the vector sum of

E_1

and

E_2

must equal zero, meaning their magnitudes must be equal.

3

Substitute values and simplify the algebraic equation

\frac{9.0 \times 10^{-9}}{x^2} = \frac{4.0 \times 10^{-9}}{(x - 0.5)^2} \implies \frac{9}{x^2} = \frac{4}{(x - 0.5)^2}

Coulomb's constant

k

and the power factor

10^{-9}

cancel from both sides.

4

Take square root on both sides to solve for x

\frac{3}{x} = \frac{2}{x - 0.5} \implies 3(x - 0.5) = 2x \implies x = 1.5\text{ m}

Taking the principal square root reduces the quadratic relation to a simple linear equation.

Key Concept

Electric Field Superposition and Zero Field Condition for Point Charges

Question 4Question →

Two point charges, $Q_1 = +3.0 \times 10^{-6}\text{ C}$ and $Q_2 = +4.0 \times 10^{-6}\text{ C}$ , are positioned in a vacuum at coordinates $(0\text{ m}, 3.0\text{ m})$ and $(3.0\text{ m}, 0\text{ m})$ respectively on a Cartesian plane. Taking the electrostatic constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , what is the magnitude of the net electric field intensity at the origin $(0,0)$ ?

1.0 \times 10^3\text{ N C}^{-1}

5.0 \times 10^3\text{ N C}^{-1}

7.0 \times 10^3\text{ N C}^{-1}

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

Show answer & explanation

Answer: $5.0 \times 10^3\text{ N C}^{-1}$

Answer

The magnitude of the net electric field intensity at the origin is

5.0 \times 10^3\text{ N C}^{-1}

.

The electric field intensity at the origin is a vector sum of the individual electric fields created by each point charge. The field due to the charge on the y-axis points downward along the y-axis with a magnitude of

3.0 \times 10^3\text{ N C}^{-1}

, while the field due to the charge on the x-axis points leftward along the x-axis with a magnitude of

4.0 \times 10^3\text{ N C}^{-1}

. Because these two fields act at right angles to each other, their vector sum magnitude is given by

\sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Step-by-Step Solution

1

Calculate the electric field intensity

E_1

at the origin due to charge

Q_1

E_1 = \frac{k |Q_1|}{r_1^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{3.0^2} = 3.0 \times 10^3\text{ N C}^{-1}

, directed along the negative y-axis.

Electric field magnitude follows Coulomb's law for field strength, and positive charges create fields directed away from themselves.

2

Calculate the electric field intensity

E_2

at the origin due to charge

Q_2

E_2 = \frac{k |Q_2|}{r_2^2} = \frac{(9.0 \times 10^9)(4.0 \times 10^{-6})}{3.0^2} = 4.0 \times 10^3\text{ N C}^{-1}

, directed along the negative x-axis.

The charge is located at

(3.0, 0)

on the x-axis, producing a field pointing toward the origin.

3

Calculate the net electric field vector magnitude at the origin

E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{(3.0 \times 10^3)^2 + (4.0 \times 10^3)^2} = 5.0 \times 10^3\text{ N C}^{-1}

.

Since the two component fields are perpendicular along orthogonal axes (x and y), their resultant is found using the Pythagorean theorem.

Key Concept

Vector addition of electric field intensities from multiple point charges

Question 5Question →

A point charge of $+5.0 \times 10^{-8}\text{ C}$ is situated in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a distance of $0.3\text{ m}$ from the charge?

Show answer & explanation

Answer: 5000

Answer

The magnitude of the electric field intensity at a distance of

0.3\text{ m}

is

5000\text{ N C}^{-1}

.

The electric field intensity

E

at a distance

r

from a point charge

q

in free space is given by

E = \frac{k |q|}{r^2}

. Substituting

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

,

q = 5.0 \times 10^{-8}\text{ C}

, and

r = 0.3\text{ m}

into the expression yields

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{(0.3)^2} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

.

Step-by-Step Solution

1

Identify the given values and formula

q = 5.0 \times 10^{-8}\text{ C}

,

r = 0.3\text{ m}

,

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

, and

E = \frac{k |q|}{r^2}

Electric field intensity surrounding a point charge depends on the charge magnitude and inversely on the square of the distance.

2

Calculate the square of the distance

r^2 = (0.3)^2 = 0.09\text{ m}^2

The inverse-square law requires using

r^2

in the denominator.

3

Substitute values and solve for field intensity

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{0.09} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

Multiplying the terms in the numerator yields

450\text{ N}\cdot\text{m}^2\text{C}^{-1}

, and dividing by

0.09\text{ m}^2

gives

5000\text{ N C}^{-1}

.

Key Concept

Electric field intensity due to a isolated point charge

Question 6Question →

Two identical insulated conducting spheres, $X$ and $Y$ , carry initial charges of $+8.0 \times 10^{-9}\text{ C}$ and $-2.0 \times 10^{-9}\text{ C}$ , respectively. They are brought into brief contact and then separated by a distance of $0.3\text{ m}$ in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force between the two spheres after separation?

9.0 \times 10^{-7}\text{ N}

1.6 \times 10^{-6}\text{ N}

2.5 \times 10^{-6}\text{ N}

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

Show answer & explanation

Answer: $9.0 \times 10^{-7}\text{ N}$

Answer

The magnitude of the electrostatic force between the spheres after separation is

9.0 \times 10^{-7}\text{ N}

.

When two identical conducting spheres touch, the total charge conserves and divides equally between them. The net charge is

(+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}

, yielding

+3.0 \times 10^{-9}\text{ C}

on each sphere. Substituting these equal charges and distance

0.3\text{ m}

into Coulomb's law gives

F = \frac{9.0 \times 10^9 \times (3.0 \times 10^{-9})^2}{0.09} = 9.0 \times 10^{-7}\text{ N}

.

Step-by-Step Solution

1

Calculate the total net charge after contact and the charge on each identical sphere.

Net charge

Q_{net} = (+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}

. Since the spheres are identical, the charge on each sphere is

q = \frac{+6.0 \times 10^{-9}\text{ C}}{2} = +3.0 \times 10^{-9}\text{ C}

.

When identical conductors touch, total charge is conserved and redistributes equally between them.

2

Apply Coulomb's Law using the redistributed charge and separation distance.

F = k \frac{q_1 q_2}{r^2} = (9.0 \times 10^9) \frac{(3.0 \times 10^{-9})(3.0 \times 10^{-9})}{(0.3)^2} = \frac{8.1 \times 10^{-7}}{0.09} = 9.0 \times 10^{-7}\text{ N}

.

Coulomb's Law calculates the magnitude of electrostatic force between two point-like charges at a given distance.

Key Concept

Charge conservation, redistribution between identical conductors, and Coulomb's Law

Estimated Time:1m 30s

Question 7Question →

A point charge of $+6.0 \times 10^{-6}\text{ C}$ experiences an attractive electrostatic force of $0.54\text{ N}$ when placed at a distance of $1.0\text{ m}$ from a second point charge in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the second charge in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 10

Answer

The magnitude of the second charge is 10 µC.

Using Coulomb's law

F = \frac{k |q_1 q_2|}{r^2}

, substituting

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

, and

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

yields

|q_2| = 1.0 \times 10^{-5}\text{ C}

, which equals

10\text{ }\mu\text{C}

.

Step-by-Step Solution

1

State Coulomb's Law formula

F = \frac{k |q_1 q_2|}{r^2}

Coulomb's Law describes the electrostatic force between two point charges.

2

Substitute given parameters into the equation

0.54 = \frac{9.0 \times 10^9 \times 6.0 \times 10^{-6} \times |q_2|}{1.0^2}

Knowns:

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

,

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

.

3

Solve for the unknown charge magnitude

q_2

|q_2| = 1.0 \times 10^{-5}\text{ C}

Rearranging yields

|q_2| = \frac{0.54}{5.4 \times 10^4} = 1.0 \times 10^{-5}\text{ C}

.

4

Convert the value from Coulombs to microcoulombs

10\text{ }\mu\text{C}

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

, so

1.0 \times 10^{-5}\text{ C} = 10\text{ }\mu\text{C}

.

Key Concept

Coulomb's Law

Question 8Question →

Two point charges, $q_1 = +1.6 \times 10^{-8}\text{ C}$ and $q_2 = +6.4 \times 10^{-8}\text{ C}$ , are fixed in a vacuum at a distance of $0.60\text{ m}$ apart. At what distance from $q_1$ along the line joining the two charges is the net electric field intensity equal to zero?

0.20\text{ m}

0.40\text{ m}

0.12\text{ m}

0.30\text{ m}

Show answer & explanation

Answer: $0.20\text{ m}$

Answer

The distance from

q_1

where the net electric field intensity is zero is

0.20\text{ m}

.

The net electric field is zero where the magnitudes of the electric fields produced by both charges are equal (

E_1 = E_2

). Setting up

\frac{k q_1}{x^2} = \frac{k q_2}{(0.60 - x)^2}

with

q_2 = 4 q_1

yields

\frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

. Taking the square root gives

\frac{1}{x} = \frac{2}{0.60 - x}

, which yields

x = 0.20\text{ m}

from

q_1

.

Step-by-Step Solution

1

Set up the condition for zero net electric field intensity.

The electric field magnitudes produced by

q_1

and

q_2

at distance

x

from

q_1

must be equal in magnitude and opposite in direction:

E_1 = E_2

.

Since both charges are positive, the point of zero net electric field must lie on the line segment connecting them.

2

Substitute the electric field formula into the equilibrium equation.

\frac{k q_1}{x^2} = \frac{k q_2}{(d - x)^2}

, where

d = 0.60\text{ m}

.

Electric field intensity due to a point charge is given by

E = \frac{k q}{r^2}

.

3

Simplify the equation by canceling common terms and substituting known charge values.

\frac{1.6 \times 10^{-8}}{x^2} = \frac{6.4 \times 10^{-8}}{(0.60 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

.

Dividing both sides by

k \times 1.6 \times 10^{-8}

reduces the numerical coefficients to simple integers.

4

Take the square root of both sides and solve for

x

.

\frac{1}{x} = \frac{2}{0.60 - x} \implies 0.60 - x = 2x \implies 3x = 0.60 \implies x = 0.20\text{ m}

.

Taking the square root removes the quadratic terms and gives a linear relation for the distance

x

from

q_1

.

Key Concept

Electric field superposition and point of zero field intensity between like point charges

Question 9Question →

A charged oil droplet of mass $3.2 \times 10^{-15}\text{ kg}$ remains stationary in a vacuum between two horizontal charged plates where there is a uniform vertical electric field of strength $2.0 \times 10^4\text{ N C}^{-1}$ . Taking $g = 10\text{ m s}^{-2}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , determine the number of excess electrons on the droplet.

Show answer & explanation

Answer: 10

Answer

The number of excess electrons on the droplet is 10.

The droplet is in mechanical equilibrium under two equal and opposite forces: the downward gravitational force

W = mg

and the upward electric force

F_e = qE

. Setting

qE = mg

gives

q = \frac{mg}{E} = 1.6 \times 10^{-18}\text{ C}

. By the quantization of charge (

q = Ne

), dividing this charge by the elementary charge

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

gives exactly 10 excess electrons.

Step-by-Step Solution

1

Calculate the gravitational force (weight) acting on the droplet.

W = mg = (3.2 \times 10^{-15}\text{ kg}) \times (10\text{ m s}^{-2}) = 3.2 \times 10^{-14}\text{ N}

.

For stationary equilibrium, weight provides the downward vertical force.

2

Apply the equilibrium condition to find the electric force.

F_e = W = 3.2 \times 10^{-14}\text{ N}

.

The net vertical force must be zero for the droplet to remain suspended.

3

Determine the charge

q

using

F_e = qE

.

q = \frac{F_e}{E} = \frac{3.2 \times 10^{-14}\text{ N}}{2.0 \times 10^4\text{ N C}^{-1}} = 1.6 \times 10^{-18}\text{ C}

.

Electric field strength relates force and charge.

4

Calculate the number of elementary charges using charge quantization

q = Ne

.

N = \frac{q}{e} = \frac{1.6 \times 10^{-18}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 10

.

Electric charge exists in discrete integer multiples of the elementary charge

e

.

Key Concept

Equilibrium between electrostatic and gravitational forces combined with charge quantization.

Question 10Question →

An insulated neutral conductor gains $5.0 \times 10^{13}$ electrons during a electrostatic charging process. Given that the magnitude of the elementary charge is $e = 1.6 \times 10^{-19}\text{ C}$ , what is the magnitude of the net charge acquired by the conductor in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 8

Answer

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

8.0\ \mu\text{C}

.

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

Q

acquired by gaining

n

electrons is given by

Q = n e

. Substituting

n = 5.0 \times 10^{13}

and

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

gives

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

. Expressed in microcoulombs,

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

.

Step-by-Step Solution

1

Apply the principle of quantization of electric charge formula

Formula

Q = n e

established

Electric charge is quantized and exists in integer multiples of the elementary charge.

2

Multiply the number of electrons by the elementary charge value

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

Calculates total electrostatic charge in base SI units.

3

Convert the value from Coulombs to microcoulombs

8.0\ \mu\text{C}

The unit

1\ \mu\text{C}

equals

10^{-6}\text{ C}

.

Key Concept

Quantization of Electric Charge

Question 11Question →

A small charged sphere of mass $2.0 \times 10^{-4}\text{ kg}$ carrying a positive charge of $+4.0 \times 10^{-8}\text{ C}$ is suspended by a light insulating string between two vertical parallel plates. When a uniform horizontal electric field of magnitude $E$ is applied between the plates, the string deflects and comes to equilibrium at an angle of $45^\circ$ to the vertical. Taking the acceleration due to gravity $g = 10\text{ m}\cdot\text{s}^{-2}$ , calculate the magnitude of the electric field intensity $E$ in $\text{N}\cdot\text{C}^{-1}$ .

Show answer & explanation

Answer: 50000

Answer

The magnitude of the electric field intensity is

50000\text{ N}\cdot\text{C}^{-1}

(or

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

).

In electrostatic equilibrium, the sphere experiences three forces: weight (

mg

) vertically downward, electrostatic force (

qE

) horizontally, and tension (

T

) along the thread at

45^\circ

to the vertical. Balancing components gives

T \sin 45^\circ = qE

and

T \cos 45^\circ = mg

. Dividing these yields

\tan 45^\circ = \frac{qE}{mg} = 1

, which gives

qE = mg

. Substituting the given values gives

E = \frac{2.0 \times 10^{-3}}{4.0 \times 10^{-8}} = 50000\text{ N}\cdot\text{C}^{-1}

.

Step-by-Step Solution

1

Calculate the weight of the charged sphere.

W = mg = (2.0 \times 10^{-4}\text{ kg})(10\text{ m}\cdot\text{s}^{-2}) = 2.0 \times 10^{-3}\text{ N}

.

The weight provides the downward vertical force in equilibrium.

2

Relate the electrostatic force to the weight using the angle of deflection.

\tan(45^\circ) = \frac{F_e}{W} \implies 1 = \frac{F_e}{2.0 \times 10^{-3}\text{ N}} \implies F_e = 2.0 \times 10^{-3}\text{ N}

.

In electrostatic equilibrium, the ratio of the horizontal force to the vertical force equals the tangent of the angle with the vertical.

3

Calculate the electric field strength

E

using

F_e = qE

.

E = \frac{F_e}{q} = \frac{2.0 \times 10^{-3}\text{ N}}{4.0 \times 10^{-8}\text{ C}} = 50000\text{ N}\cdot\text{C}^{-1}

.

The electric field intensity is the electric force per unit charge.

Key Concept

Equilibrium of a charged body in a uniform electric field

Estimated Time:2m 0s

Question 12Question →

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 13Question →

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 14Question →

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 15Question →

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 16Question →

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 17Question →

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 18Question →

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 19Question →

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 20Question →

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

Electrostatics and Electric Charges

Answer

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