A slab of material of dielectric constant k, having the same area as the plate of a parallel plate capacitor but has a thickness \( \displaystyle\frac{3}{5} \) of the distance between the plates d, is inserted between the plates of the capacitor. When there is air between the plates, the potential difference is \( V_0 \). What is the potential difference between the plates of the capacitor after inserting the dielectric?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\left(\frac{2k+3}{5k}\right)V_0 \)
The capacitor has been charged to \( V_0 \) and is then isolated, so the charge on the plates, and with it the field in the air gap, stays the same after the slab is inserted. With air, the field is \( E_0 = V_0/d \). Inside the dielectric the field is reduced to \( E_0/k \).
The slab has thickness \( t = \frac{3d}{5} \), which leaves an air gap of \( d - t = \frac{2d}{5} \). The potential difference is the sum of the drops across the two parts:
\( \displaystyle V = E_0\,(d-t) + \frac{E_0}{k}\,t \)
\( \displaystyle V = \frac{V_0}{d}\left(\frac{2d}{5}\right) + \frac{V_0}{kd}\left(\frac{3d}{5}\right) \)
\( \displaystyle V = V_0\left(\frac{2}{5}+\frac{3}{5k}\right) = \left(\frac{2k+3}{5k}\right)V_0 \)
Shortcut: Test the limits. For \( k = 1 \) (no real dielectric) V must equal \( V_0 \); Options (2) and (3) both give 1, but Options (1) and (4) do not. For \( k\to\infty \) the slab behaves like a conductor and the potential difference falls to \( \frac{2}{5}V_0 \), the drop across the remaining air gap. Option (2) gives \( \frac{2}{5} \), while Option (3) tends to \( \frac{5}{2} \), an increase, which cannot happen when a dielectric is inserted.