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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

A parallel pl...

Question

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by $Q_{0}, V_{0}, E_{0}$ , and $U_{0}$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ are related to the previous ones are :

Q > Q_{0}

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V > V_{0}

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E > E_{0}

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U < U_{0}

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Solution

The correct option is A $Q > Q_{0}$
As the battery till in connection so potential will remained unchanged. i,e $V = V_{0}$
$E_{0} = \frac{V_{0}}{d}$ and $E = \frac{V}{d} = \frac{V_{0}}{d} \Rightarrow E = E_{0}$ where $d =$ separation of plates.
Let initial capacitance $= C_{0}$
When dielectric slab with dielectric constant K is inserted, the capacitance becomes $C = K C_{0}$ .
Thus, $Q_{0} = C_{0} V_{0}$ and $Q = C V = K C_{0} V_{0} = K Q_{0} \Rightarrow Q > Q_{0}$ as $(K > 1)$
Also, $U_{0} = (1 / 2) C_{0} V_{0}^{2}$ and $U = (1 / 2) C V^{2} = (1 / 2) K C_{0} . V_{0} = K U_{0}$
Hence, $U > U_{0}$

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