Question

How many electrons should be removed from an initially uncharged spherical conductor of radius 0.200 m to produce a potential of 5.50 kv at the surface?

Answer 1

To calculate the number of electrons from spherical conductor first we use the formula,

$V= k\frac{q}{r}$

Here, k is a constant with a value of $8.99 \times 10^9 N.m^{2} /C^2$ , q is the charge and r is the radius and its value of 0.200 m.

Substituting these value in above formula, we get  

$5.50 kV =\frac{8.99 \times 10^9 N.m^{2} /C^2\times q}{0.200 m}$

or                $q= \frac{5.50\times10^3 V\times0.200 m}{8.99 \times 10^9 N.m^{2} /C^2}$

$q=1.2\times 10^{-7} C$

Now number of electron,

$N= \frac{1.2\times 10^{-7} C}{1.6\times 10^{-19}C/e }$

$N=7.5\times 10^{11} electrons$

Hence, the number of electrons to be removed from conductor would be $7.5\times 10^{11} electrons$

                     

Answer 2

Solution:

Potential is defined as

V = kq/r  

=> q= Vr/k  

= 5500 * 0.2 / 9 x 10^9 C  

= (110/9) x 10^-7  

If n = no. of electrons to be removed

So,  

n x 1.6 x 10^-19  

Therefore,

n=v/q

=> n = (110/14.4) x 10^12 electrons  

= 7.6382 x 10^12 electrons.  

This will leave an equal amount of positive charge in protons and produce the give potential on the surface of the sphere.