2. Electric Potential and Capacitance
easy

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

$3{Q_0},\;3{V_0},\;3{E_0}$

B

${Q_0},\;3{V_0},\;3{E_0}$

C

${Q_0},\;\frac{{{V_0}}}{3},\;3{E_0}$

D

${Q_0},\;\frac{{{V_0}}}{3},\;\frac{{{E_0}}}{3}$

Solution

(d) 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 $E' = \frac{{{E_0} \times 3}}{9} = \frac{{{E_0}}}{3}$

Standard 12
Physics

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