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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Two identical metal spheres possess $+ 60 C$ and $- 20 C$ of charges. They are brought in contact and then separated by $10 c m$ . The force between them is:

Two charges when kept at a distance of $1 m$ apart in vacuum have some force of repulsion. If the force of repulsion between these two charges be same, when placed in an oil of dielectric constant $4$ , the distance of separation is :

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Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium seperation between the balls now become

Two identical metal spheres possess +60 C and −20 C of charges. They are brought in contact and then separated by 10 cm. The force between them is:

Two charges when kept at a distance of 1m apart in vacuum have some force of repulsion. If the force of repulsion between these two charges be same, when placed in an oil of dielectric constant 4, the distance of separation is :