Question

Find the capacitance of a system of two identical metal balls of radius a If the distance between there centers is equal to b ,. with b>>a. the system is located in a uniform dielectric with permittivity K​

Answer 1

$\huge\mathtt\pink{A}\mathtt\red{N}\mathtt\blue{S}\mathtt\green{W}\mathtt\purple{E}\mathtt\green{R}\: 2π\epsilon_o K a$

Explanation:

As we know that C=Q/∆V

And dv=-E.dr

Hence at distance x Electric field and Potential relationship can be written as

$\delta v = \int e.dr$

$v = \int_{ a}^{b - a} e.dr$

$v = \int_{ a}^{b - a} ( \frac{kq}{K {x}^{2}} + \frac{kq}{(K( {b - x})^{2} }) dx$

$v = \int_{ a}^{b - a} ( \frac{ - 1}{ {x}} + \frac{1}{( {b - x})})/K$

$v =kq ( \frac{ - 1}{ {b - a}} + \frac{1}{( {b - (b - a)})} - ( \frac{ - 1}{a} + \frac{1}{b - a} )/K$

$v =kq ( \frac { 2}{ {aK}} - \frac{2}{( K{b - a})})$

Now put it in formula of capacitance

$c = \frac{q}{v}$

$c = \frac{qK}{kq( \frac{ 2}{ {a}} - \frac{2}{( {b - a})})}$

$c = \frac{4\pi \epsilon_o K(ab - a)}{2b - 4a}$

$c = {2\pi \epsilon_oKa}$

NoTE:-

Capital K stands for dielectric constant .

Value of Electric field become Kth times by adding dielectric that's why K is added in denominator everywhere