Define capacitance and state its unit - Leaving Cert Physics - Question c - 2018

The energy (E) stored in a capacitor can be calculated using the formula:

$E = \frac{1}{2} C V^2$

First, we calculate the capacitance (C). Given a charge of 0.11 C and a potential difference of 4.0 kV (or 4000 V):

$C = \frac{Q}{V} = \frac{0.11}{4000} = 2.75 \times 10^{-5} \text{ F}$

Then, substituting this value into the energy formula:

$E = \frac{1}{2} (2.75 \times 10^{-5}) (4000)^2 = 220 \text{ J}$

The net charge of the capacitor when it stores this energy is 0.11 C.

The average current (I) can be calculated using the formula:

$I = \frac{Q}{t}$

Given that the capacitor discharges a charge Q = 0.11 C in a time t = 15 ms (or 0.015 s):

$I = \frac{0.11}{0.015} = 7.33 \text{ A}$

The electric field between the charged plates of a parallel plate capacitor can be represented with plates labeled as positive (+) and negative (-). The field lines should be drawn as straight lines emanating from the positive plate, pointing towards the negative plate, indicating the direction of the electric field.