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CGP EDU Academic Team
Published on: September 12, 2026
A capacitor when filled with a dielectric $K = 3$ has charge $Q_0$ , voltage $V_0$ and field $E_0$ . If the dielectric is replaced with another one having K = 9 the new values of charge, voltage and field will be respectively
Text Solution
Verified by ExpertsThe correct answer is:
D
When there is no battery, charge remains same while potential difference and electric field decreases
i.e. $Q' = Q_0, \quad V' = \frac{V_0 \times 3}{9} = \frac{V_0}{3}$ and $E' = \frac{E_0 \times 3}{9} = \frac{E_0}{3}$
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