Question

The distance between the plates of a charged parallel plate capacitor is $5\text{cm}$ and the electric field inside the plates is $200\text{V/cm}$. An uncharged metal plate of same length and width $2\text{cm}$ is inserted between the capacitor. The voltage across capacitor after insertion of the plate is

• A. $0$
• B. $400\text{V}$
• C. $600\text{V}$
• D. $100\text{V}$

Answer 1

The correct option is C $600\text{V}$

Initially capacitor will be as shown below.


$E=200\frac{\text{V}}{\text{cm}};d=5\text{cm}$


After insertion of metallic plate inside the $\text{PPC}$


As we know that electric field inside the conductor will be zero.

Everywhere else in the region between
the plates, $E$ remains $200\frac{\text{V}}{\text{cm}}$

We know, $\Delta V=E\Delta d$

$\Rightarrow \Delta V=200\frac{\text{V}}{\text{cm}}\times (5-2)\text{cm}=600\text{V}$

Hence, option $\left(c\right)$ is correct.

Key concept- If the electric field is uniform, potential drop is electric field times distance moved along the direction of the field.