Respuesta :
To calculate the number of electrons from spherical conductor first we use the formula,
[tex]V= k\frac{q}{r}[/tex]
Here, k is a constant with a value of [tex]8.99 \times 10^9 N.m^{2} /C^2[/tex] , q is the charge and r is the radius and its value of 0.200 m.
Substituting these value in above formula, we get
[tex]5.50 kV =\frac{8.99 \times 10^9 N.m^{2} /C^2\times q}{0.200 m}[/tex]
or [tex]q= \frac{5.50\times10^3 V\times0.200 m}{8.99 \times 10^9 N.m^{2} /C^2}[/tex]
[tex]q=1.2\times 10^{-7} C[/tex]
Now number of electron,
[tex]N= \frac{1.2\times 10^{-7} C}{1.6\times 10^{-19}C/e }[/tex]
[tex]N=7.5\times 10^{11} electrons[/tex]
Hence, the number of electrons to be removed from conductor would be [tex]7.5\times 10^{11} electrons[/tex]
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.
