Question

A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is.

• A. $\frac{\varepsilon_{o}A}{d - \frac{b}{2}}$
• B. $\frac{\varepsilon_{o}A}{2d}$
• C. $\frac{\varepsilon_{o}A}{d - b}$
• D. $\frac{\varepsilon_{o}A}{d - \frac{b}{2}}$

Answer 1

Answer: (C)

$\frac{\varepsilon_{o}A}{d - b}$

Answer 2

: As capacitance$C_{o} = \frac{\varepsilon_{o}A}{d}$

$\therefore$ After inserting copper plate

$C = \frac{\varepsilon_{o}A}{d - b}$