alleryStart learning →A parallel-plate capacitor has capacitance $C$. If BOTH the plate area and the plate separation are doubled, the new capacitance is
A$C$
B$4C$
C$2C$
D$C/2$
Answer & Solution
Correct answer: A. $C$
1. For a parallel-plate capacitor $C = \dfrac{\varepsilon_0 A}{d}$.
2. Doubling $A$ doubles the numerator; doubling $d$ doubles the denominator.
3. The two changes cancel: $C' = \dfrac{\varepsilon_0(2A)}{2d} = \dfrac{\varepsilon_0 A}{d} = C$ (unchanged).
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Long Answer Q.18 / Evaluation Q.13_
Keep practising Electromagnetism
Browse all Electromagnetism questions for Tamil Nadu HSC (Class 12), or explore more Physics topics.
Related questions
On the axial line of a short dipole the electric field is $E_a$, and on the equatorial linThe energy stored in a capacitor of capacitance $C$ charged to voltage $V$ (holding chargeBy Gauss's law, the total electric flux through a closed surface enclosing a net charge $QAn equipotential surface is one on whichThe relation between the electric field and the electrostatic potential isIf the voltage across a capacitor is increased from $V$ to $2V$, thenPoints A and B are at potentials $+7\ V$ and $-4\ V$. The work done in moving 50 electronsTwo objects rubbed together acquire a charge of $50\ nC$ each. The number of electrons tra