Which of the following is correct for a parallel plate capacitor of separation 'd' when a slab of material of dielectric constant \( k \) has the same area as the plate of capacitor but has a thickness d/2 is inserted between the plates?
A. Its capacitance decreases
B. Its capacitance increases
C. \( \displaystyle C=C_0\left(\frac{2k}{k+1}\right) \)
D. \( \displaystyle C=C_0\left(\frac{k+1}{2k}\right) \)
E. Charge remains same.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → B, C and E Only
A slab of thickness t and dielectric constant k filling the full plate area makes the capacitor equivalent to two capacitors in series: an air gap of thickness \( d - t \) and the slab of thickness t. The capacitance with the slab is \( C = \frac{\varepsilon_0A}{(d - t) + t/k} \), and without it \( C_0 = \frac{\varepsilon_0A}{d} \).
For \( t = d/2 \):
\( \displaystyle C = \frac{\varepsilon_0A}{\frac{d}{2} + \frac{d}{2k}} = \frac{2k\,\varepsilon_0A}{(k+1)\,d} \)
\( \displaystyle C = C_0\left(\frac{2k}{k+1}\right) \)
(C) \( C = C_0\left(\frac{2k}{k+1}\right) \) — Correct. This is the result just derived.
(D) \( C = C_0\left(\frac{k+1}{2k}\right) \) — Incorrect. It is the inverse of the correct factor.
(B) Its capacitance increases — Correct. Since \( k \gt 1 \), we have \( 2k \gt k + 1 \), so the factor \( \frac{2k}{k+1} \) is greater than 1.
(A) Its capacitance decreases — Incorrect. The capacitance increases.
(E) Charge remains same — Correct for a charged capacitor that is isolated: inserting the slab changes the capacitance and the potential difference but not the charge on the plates.
Only Option (4) has both B and C.