Assertion (A): If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant is made three times, the capacitance becomes 6 times.
Reason (R): The capacitance of a capacitor does not depend upon the nature of the dielectric material.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) is correct but (R) is incorrect.
The capacitance of a parallel plate capacitor with a dielectric filling the gap is \( C=\varepsilon_0KA/d \). It is directly proportional to the dielectric constant K and inversely proportional to the plate separation d.
With \( K'=3K \) and \( d'=d/2 \), the new capacitance is:
\( \begin{aligned} C' &= \frac{\varepsilon_0\,(3K)\,A}{d/2} \\ &= 6\,\frac{\varepsilon_0KA}{d} \\ &= 6C \end{aligned} \)
Assertion (A) If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant is made three times, the capacitance becomes 6 times. — True. Tripling K multiplies the capacitance by 3 and halving d multiplies it by 2, so it becomes \( 3\times2=6 \) times.
Reason (R) The capacitance of a capacitor does not depend upon the nature of the dielectric material. — False. The dielectric constant K is a property of the material between the plates. A dielectric that fills the gap raises the capacitance by the factor K from its vacuum value.
NCERT (2026-27):
Effect of Dielectric on Capacitance – “… The capacitance C, with dielectric between the plates, is then
\( \displaystyle C=\frac{Q}{V}=\frac{\varepsilon_0KA}{d} \)\( \qquad (2.51) \)”
Effect of Dielectric on Capacitance – “…
\( \displaystyle K=\frac{C}{C_0} \qquad (2.54) \)
Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.”