Question

Two identical metallic spheres having charges +4q and -2q are placed with their centres r distance apart. Force of attraction between the spheres is F. If the two spheres are brought in contact and then placed at the same distance r apart, the force between them is
A) F
B) F/2
C) F/4
D) F/8

Answer 1

Answer:

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Answer 2

Given:

Two identical metallic spheres having charges +4q and -2q are placed with their centres r distance apart. Force of attraction between the spheres is F.

To find:

New force when the spheres are touched and placed at same separation?

Calculation:

• Let initial separation between charges be r, so initial force will be :

$F = \dfrac{k(4q)(2q)}{ {r}^{2} }$

$\implies \: F = \dfrac{8k {q}^{2} }{ {r}^{2} }$

• Now, when the spheres are touched, the new charge in each sphere will be average value = [4+(-2)]q/2 = 2q/2 = 1q

So, new force is :

$F_{2} = \dfrac{k(q)(q)}{ {r}^{2} }$

$\implies F_{2}= \dfrac{k {q}^{2} }{ {r}^{2} }$

• Multiply numerator and denominator by 8.

$\implies F_{2} = \dfrac{1}{8} \times \dfrac{8k {q}^{2} }{ {r}^{2} }$

$\implies F_{2} = \dfrac{1}{8} \times F$

$\implies F_{2} = \dfrac{F}{8}$

So, option D) is correct ✔️