Question

The radii of two metallic spheres are 5cm and 10cm and both carry an equal charge of 75 $\mu C$. If the two spheres are shorted then the charge will be transferred.
A. 25 $\mu C$ from smaller to bigger
B. 25 $\mu C$ from bigger to smaller
C. 50 $\mu C$ from smaller to bigger
D. 50 $\mu C$ from bigger to smaller

Answer 1

Hint: When the two metallic spheres are shorted then the charges will move to equalize the potential on each of them. Since there is no loss of charge in transfer, find the difference between charges in each sphere before and after the shorting.

Complete answer:
Let A and B be the metallic spheres with radii ${r_1} = 5cm,{r_2} = 10cm = 2{r_1}$ respectively with the same charge $Q = 75\mu C$. Let the charges on each of the spheres after they have been shorted be ${Q_1},{Q_2}$ respectively.
After both of them have been shorted, potential will balance and be as follows:
Electric Potential of A = Electric Potential of B
$ \Rightarrow \dfrac{{K{Q_1}}}{{{r_1}}} = \dfrac{{K{Q_2}}}{{{r_2}}}$
Therefore,
$\Rightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{r_1}}}{{{r_2}}} = \dfrac{1}{2}$
$\Rightarrow {Q_2} = 2{Q_1}$
If we assume no loss of charges before and after the shorting, then we have the relations,
$\Rightarrow {Q_1} + {Q_2} = 75 + 75 = 150\mu C$
$\Rightarrow 3{Q_1} = 150$
$\Rightarrow {Q_1} = 50\mu C;{Q_2} = 2{Q_1} = 100\mu C$
Now if we compare the charges before and after the shorting, we see that 25 $\mu C$ of charge has moved from sphere A of radius 5cm to sphere B of radius 10cm. Option A is the correct one.

Note: For such problems it is important to remember the charges flow to balance the potential from one point to another. First right down the given conditions and relations and then use the concept of balancing potential.