Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force $F$ , when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

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Answer 1

$Step 1: Calculation of initial force$

Let $B$ and $C$ have charge $Q$ each initially and are seperated by $r$ distance.

From Coulomb's law, force between them $F = \frac{K Q ^{2}}{r ^{2}}$ $. . . . (1)$

$Step 2: Distribution of charges$

When an identical uncharged spherical conductor $A$ is brought in contact with charged sphere $B$ .
By symmetry, the total charge will be shared equally among them, as both have equal radii.
$∴$ Charge on $A$ and $B$ $= \frac{Q}{2}$

Now, when the conductor $A$ is brought in contact with $C$ .
They will also share equal charge among themselves, as both have equal radii.
$∴$ Charge on $A$ and $C$ $= \frac{Q + \frac{Q}{2}}{2} = \frac{3 Q}{4}$

$Step 3: Calculation of new force$

New coulomb force between $B$ and $C$ :

$F^{'} = \frac{K \frac{Q}{2} \times \frac{3 Q}{4}}{r ^{2}} = \frac{3}{8} \frac{K Q ^{2}}{r ^{2}}$

From equation $(1)$ , we get: $F^{'} = \frac{3}{8} F$

Hence, Option D is correct

Answer 2

Step 1: Calculation of initial force

Let

B

and

C

have charge

Q

each initially and are seperated by

r

distance.

From Coulomb's law, force between them

F = \frac{K Q ^{2}}{r ^{2}}

. . . . (1)

Step 2: Distribution of charges

When an identical uncharged spherical conductor

A

is brought in contact with charged sphere

B

.

By symmetry, the total charge will be shared equally among them, as both have equal radii.

∴

Charge on

A

and

B

= \frac{Q}{2}

Now, when the conductor

A

is brought in contact with

C

.

They will also share equal charge among themselves, as both have equal radii.

∴

Charge on

A

and

C

= \frac{Q + \frac{Q}{2}}{2} = \frac{3 Q}{4}

Step 3: Calculation of new force

New coulomb force between

B

and

C

:

F^{'} = \frac{K \frac{Q}{2} \times \frac{3 Q}{4}}{r ^{2}} = \frac{3}{8} \frac{K Q ^{2}}{r ^{2}}

From equation

(1)

, we get:

F^{'} = \frac{3}{8} F

Hence, Option D is correct

$\frac{F}{4}$

$\frac{3 F}{4}$

$\frac{F}{8}$

$\frac{3 F}{8}$

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