Practicing Success
The plates of a parallel plate capacitor in vacuum are 5.0 mm apart. A potential difference of 10.0 kV is applied. The magnitude of the surface charge density on each plate is
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1.77 x 10–5 Cm–2 3.54 x 10–5 Cm–2 1.77 x 10–3 Cm–2 3.54 x 10–3 Cm–2 |
1.77 x 10–5 Cm–2 |
$ \text{ Charge on plates of capacitor is } Q = CV , C =\frac{\epsilon_0 A}{d}$ $ \text{Surface Charge density } \sigma = \frac{Q}{A} = \frac{CV}{A} = \frac{\epsilon_0 V}{d}$ $ \sigma = \frac{8.85\times 10^{-12}\times 10^4}{5\times 10^{-3}} = 1.77 \times 10^{-5} C/m^2$ |