Electrostatics
Intermediate
Capacitor
Source: High school physics (Chinese)
Problem
A capacitor charged with charge $Q$ has a potential difference $U$ across its plates. When the charge is increased by $4.0 \times 10^{-8}$ C, the potential difference increases by $20$ V.
Find the capacitance of the capacitor.
$C = 2.0 \times 10^{-9}$ F $= 2.0$ nF.
For a capacitor, $Q = CU$, so any change satisfies $\Delta Q = C \, \Delta U$ (capacitance is a constant determined by geometry, independent of charge state). Therefore:
$$C = \dfrac{\Delta Q}{\Delta U} = \dfrac{4.0 \times 10^{-8}\ \mathrm{C}}{20\ \mathrm{V}} = 2.0 \times 10^{-9}\ \mathrm{F}.$$