Figure shows a parallel plate capacitor with plate area $A$ and plate separation $d$ . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant $K$ is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

$\frac{V}{d}$
$\frac{K V}{d}$
$\frac{V}{d - t}$
$\frac{ε A V}{d}$

Question

Figure shows a parallel plate capacitor with plate area $A$ and plate separation $d$ . A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant $K$ is placed in between the plates of the capacitor as shown.
Now, answer the following questions based on above information. The electric field in the gaps between the plates and the dielectric slab will is :

$\frac{V}{d}$
$\frac{K V}{d}$
$\frac{V}{d - t}$
$\frac{ε A V}{d}$

Toppr Admin · Accepted Answer

The gap between the plates and the dielectric field is filled with air-vacuum- -xA0-So electric field here is same as electric field with vacuum between the parallel plate capacitor- i-e- Eo-V-d

The gap between the plates and the dielectric field is filled with air/vacuum. So electric field here is same as electric field with vacuum between the parallel plate capacitor, i,e, Eo=V/d

The gap between the plates and the dielectric field is filled with air/vacuum. So electric field here is same as electric field with vacuum between the parallel plate capacitor, i,e, $E_{o} = V / d$