A parallel plate capacitor is charged by a battery. After some ti… - Vidyadip

Question

A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. What change, in any will take place in
i. charge on the plates
ii. electric field intensity between the plates
iii. the capacitance of the capacitor,
iv. a potential difference between the plates and
v. the energy stored in the capacitor? Justify your answer in each case. OR

✓

Answer

i. The charge $q _0$ on the capacitor plates remains the same because the battery has been disconnected, before placing the dielectric slab.
ii. The surface charges induced on the dielectric slab reduce electric field intensity to a new value given by $E=\frac{E_0}{\kappa}$
iii. The reduction in the electric field induces the potential difference $V=E d=\frac{E_0 d}{k}=\frac{V_0}{k}$
iv. Due to the decrease in p.d., the capacitance increases k times $C=\frac{q_0}{V}=\frac{q_0}{V_0 / k }= K \frac{q_0}{V_0}=\kappa C _0$ v. Energy stored decreases by a factor of $\kappa$ :
\[
U=\frac{1}{2} C V^2=\frac{1}{2}\left(\kappa C_0\right)\left(\frac{V_0}{K}\right)^2=\frac{1}{\kappa} \cdot \frac{1}{2} C_0 V_0^2=\frac{U_0}{\kappa}
\]

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