Question

A charged capacitor has energy $U_{0} = 2 \times 10^{- 4} J$ and $C_{0} = 8 \times 10^{- 8} J$ , in the absence of a dielectric. When the potential difference is kept constant by keeping the capacitor connected to the supply, assuming $ε_{r} = 150$

(i) find the new energy stored when the dielectric slab is completely filled with the capacitor. Account for this change in energy

(ii) how much excess charge flows from the supply

(iii) find the ratio of the charge on the plate of the capacitor and the charge induced on the dielectric

Accepted Answer

$(i)$ We are given a charged capacitor with energy, $U_{0} = 2 \times 10^{- 4} J$

Capacitance, $C_{0} = 8 \times 10^{- 8} J$

Dielectric constant, $ε_{r} = 150$

Capacitance with dielectric is increases by $C' = ε_{r} C_{0}$ .

The potential difference $V$ is kept constant by keeping it connected to the supply.

Energy stored in capacitor is

$U = \frac{1}{2} C' V^{2} = \frac{1}{2} (ε_{r} C_{0}) V^{2} U = ε_{r} (\frac{1}{2} C_{0} V^{2}) = ε_{r} U_{0}$

$U = (2 \times 10^{- 4}) (150) = 0 .33 J$

The final energy after adding dielectric is greater than the initial energy.

The additional energy is $∆ U = U_{0} (ε_{r} - 1) > 0$ .

The external supply provides this extra energy to the circuit.

$(i i)$ Charge stored in capacitor is $q = C V = ε_{r} C_{0} V$

Initial energy stored in capacitor is $U_{0} = \frac{1}{2} C_{0} V^{2}$ .

$V^{2} = \frac{2 \times 2 \times 10^{- 4}}{8 \times 10^{- 8}} V = \sqrt{\frac{10^{4}}{4}} = 50 V$

Now, excess charge is $∆ q = (∆ C) V$ .

where $∆ C$ is the change in capacitance of the system.

$∆ q = (C - C_{0}) V = (ε_{r} - 1) C_{0} V = 149 \times 4 \times 10^{- 8}$

The excess charge flows from the supply is $5 .96 \times 10^{- 6} C$ .

$(i i i)$ The ratio of $\frac{charge on the plate of capacitor}{charge induced on the dielectric} = \frac{q_{f}}{q_{ind}}$

$\frac{q_{ind}}{q_{f}} = \frac{∆ q}{q} = 1 - \frac{1}{ε_{r}}$

$q_{ind} = q_{f} (1 - \frac{1}{ε_{r}}) = \frac{149}{150} q_{f}$

Hence, the ratio of $\frac{q_{f}}{q_{induced}}$ is $\frac{150}{149}$ .

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