Question

Two capacitors are joined in series as shown in figure. The area of each plate is A. The equivalent of the combination is –

• A. (a) $\frac{\varepsilon_{0}A}{d_{1} - d_{2}}$
• A. (b)$\frac{\varepsilon_{0}A}{a - b}$
• A. (c) $\varepsilon_{0}A\left( \frac{1}{a} - \frac{1}{b} \right)$
• A. (d)

Answer 1

(b)

When two capacitors are in series

$\frac { 1 } { \mathrm { C } _ { \mathrm { eq } } }$= =

\ C_eq =

=

Now in given arrangement

Capacitors are in series

& sum of separations = a – b

\ C_eq =