Question

Kindly Answer this question.
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Answer 1

hey there

refer to attachment

Answer 2

Question:

Two conductors carrying equal and opposite charges create a non - uniform field as shown in the figure given below. What will be the capacity of this capacitor if the field along Y - axis varies as $E = \frac{Q}{ {\epsilon}_{0}} \left[1 + B {Y}^{2} \right]$, where B is a constant.

Answer:

$\large\boxed{\sf{C = \frac{3 {\epsilon}_{0} A}{a(3 + B {a}^{2}) } }}$

Explanation:

It's been given that:

$E = \frac{Q}{ {\epsilon}_{0}} \left[1 + B {Y}^{2} \right]$

But, we know that:

$E = - \frac{dV}{dr}$

Therefore, we get,

$-\frac{dV}{dr} = \frac{Q}{ {\epsilon}_{0}} \left[1 + B {Y}^{2} \right]$

Further solving, we get:

Refer to the attachment now.

From the attached solution, we get that,

$\frac{{V}_{A} - V_{B}}{Q} =\frac{1}{ \frac{3 {\epsilon}_{0} A}{a(3 + B {a}^{2}) } }$

But, we know that,

$\dfrac{Q}{V}=C$

Hence, we get the capacity as:

$\large\boxed{\red{C = \frac{3 {\epsilon}_{0} A}{a(3 + B {a}^{2}) } }}$