Question

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

• A. $3Q_{0},\mspace{6mu} 3V_{0},\mspace{6mu} 3E_{0}$
• B. $Q_{0},\mspace{6mu} 3V_{0},\mspace{6mu} 3E_{0}$
• C. $Q_{0},\mspace{6mu}\frac{V_{0}}{3},\mspace{6mu} 3E_{0}$
• D. $Q_{0},\mspace{6mu}\frac{V_{0}}{3},\mspace{6mu}\frac{E_{0}}{3}$

Answer 1

Answer: (D)

$Q_{0},\mspace{6mu}\frac{V_{0}}{3},\mspace{6mu}\frac{E_{0}}{3}$

Answer 2

When there is no battery, charge remains same while potential difference and electric field decreases

i.e. $Q' = Q_{0},V' = \frac{V_{0} \times 3}{9} = \frac{V_{0}}{3}$and $E' = \frac{E_{0} \times 3}{9} = \frac{E_{0}}{3}$