Two identical isolated metal spheres, X and Y, carry initial charges of +q and -3q respect… — JAMB UTME

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 $F$ ?

$\dfrac{35}{12}F$ Answer
B
$\dfrac{5}{4}F$
C
$\dfrac{35}{48}F$
D
$\dfrac{35}{4}F$

Answer

The magnitude of the net electrostatic force acting on sphere

Z

\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

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.

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.

Determine the forces exerted on sphere

Z

at the midpoint.

Separation distance from

Z

to both

X

and

Y

\frac{r}{2}

. Force from

X

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

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

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

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Key Concept

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