Question

A parallel-plate capacitor of capacitance C consists of two identical circular plates each of radius r, seperated by a distance d. The capacitance of a capacitor having plates of radius r/3 and seperated by a distance d/3 would be
A. 3C
B. C
C. C/3
D. C/9​

Answer 1

Given:

A parallel-plate capacitor of capacitance C consists of two identical circular plates each of radius r, seperated by a distance d.

To find:

New capacitance when plates have radius r/3 and seperated by a distance d/3 ?

Calculation:

Initially, Capacitance of Parallel Plate Capacitor will be :

$\rm \: C = \dfrac{A \epsilon_{0} }{d}$

$\rm \implies\: C = \dfrac{\pi {r}^{2} \epsilon_{0} }{d}$

Now, new radius is r/3 and distance between plates is d/3.

$\rm \implies\: C_{new} = \dfrac{\pi { \bigg( \dfrac{r}{3} \bigg) }^{2} \epsilon_{0} }{ \bigg(\dfrac{d}{3} \bigg) }$

$\rm \implies\: C_{new} = \dfrac{3\pi {r}^{2} \epsilon_{0} }{ 9d }$

$\rm \implies\: C_{new} = \dfrac{1}{3} \times \dfrac{\pi {r}^{2} \epsilon_{0} }{ d }$

$\rm \implies\: C_{new} = \dfrac{C}{3}$

So, new capacitance is C/3.

Answer 2

Explanation:

$Given:</p><p></p><p>A parallel-plate capacitor of capacitance C consists of two identical circular plates each of radius r, seperated by a distance d.</p><p></p><p>To find:</p><p></p><p>New capacitance when plates have radius r/3 and seperated by a distance d/3 ?</p><p></p><p>Calculation:</p><p></p><p>Initially, Capacitance of Parallel Plate Capacitor will be :</p><p></p><p>\rm \: C = \dfrac{A \epsilon_{0} }{d}C=dAϵ0</p><p></p><p>\rm \implies\: C = \dfrac{\pi {r}^{2} \epsilon_{0} }{d}⟹C=dπr2ϵ0</p><p></p><p>Now, new radius is r/3 and distance between plates is d/3.</p><p></p><p>\rm \implies\: C_{new} = \dfrac{\pi { \bigg( \dfrac{r}{3} \bigg) }^{2} \epsilon_{0} }{ \bigg(\dfrac{d}{3} \bigg) }⟹Cnew=(3d)π(3r)2ϵ0</p><p></p><p>\rm \implies\: C_{new} = \dfrac{3\pi {r}^{2} \epsilon_{0} }{ 9d }⟹Cnew=9d3πr2ϵ0</p><p></p><p>\rm \implies\: C_{new} = \dfrac{1}{3} \times \dfrac{\pi {r}^{2} \epsilon_{0} }{ d }⟹Cnew=31×dπr2ϵ0</p><p></p><p>\rm \implies\: C_{new} = \dfrac{C}{3}⟹Cnew=3C</p><p></p><p>So, new capacitance is C/3.</p><p></p><p>$