Question

The distance between the plates of a charged Parallel plate capacitor is $5\text{mm}$ and the electric field inside the plates is $40\text{V/mm}$. An uncharged metal plate of width $1\text{mm}$ is fully immersed into the capacitor. The length of the metal bar is the same as that of the plates of the capacitor. The voltage across the capacitor after insertion of the bar is

• A. $320\text{V}$
• B. $40\text{V}$
  
• C. $160\text{V}$
• D. $80\text{V}$

Answer 1

The correct option is C $160\text{V}$

Before insertion of the conductor, capacitor will be as shown below.


So, potential difference across plate will be

$V=Ed=40\times 5=200\text{V}$

After insertion of the conductor,


As we know that electric field inside the conductor is zero, and everywhere else in the region between the plates $E$ remains $40\text{V/mm}$.

So, potential difference will be

$V^{\prime }=E\Delta d=40\text{V/mm}\times (5-1)\text{mm}$

$\therefore V^{\prime }=160\text{V}$

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

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