Question

The radii of two metallic spheres are 5cm & 10cm and both carry equal charges of 75 micro coulomb . If the two spheres are shorted then charge will be transferred :-
(A) 25 m.c. from smaller to bigger
(B) 25 m.c. from bigger to smaller
(C) 50 m.c. from smaller to bigger
(D) 50 m.c. from bigger to smaller

Answer 1

Answer: The charge will be transferred 25 micro Coulomb from smaller to bigger

Explanation:

Given that,

Equal charges of the metallic spheres

$q_{1} = q_{2} = 75\mu C$

The Radius of the two metallic spheres is

$r_{1} = 5 cm$

$r_{2} = 10 cm$

If the two spheres are shorted that means charges locked at equi potential.

Suppose after the charge transfer, final charges be $q_{1}$ and $q_{2}$

Smaller sphere is at higher potential and bigger sphere is at lower potential.

So, the charge transferred from smaller sphere to higher sphere

The potential of the spheres would be same when they are shorted.

$V_{1} = V_{2}$

$\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}}{r_{1}} = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{2}}{r_{2}}$

$\dfrac{q_{1}}{5} = \dfrac{q_{2}}{10}$

$q_{2} = 2q_{1}$......(I)

Now, the net charge will be the same at  starting and end

So, $q_{1} + q_{2} = 150\mu C$....(II)

Now, put the value of $q_{2}$ in equation(II)

$q_{1} + 2q_{1} 150\mu C$

$3q_{1} 150\mu C$

$q_{1} = 50\mu C$

Now, put the value of $q_{3}$ in equation (I)

$q_{2} = 2\times50\mu c$

$q_{2} = 100\mu C$

Hence, the charge will be transferred 25 micro Coulomb from smaller to bigger.