Question

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?

Short answer

Solution: The capacitance of the parallel plate capacitor is approximately \(8.33 \times 10^{-11} \,\mathrm{F}\).

Step-by-step solution

Convert given quantities to standard units

Firstly, we convert given quantities to standard units. The charge given in microcoulombs should be converted to coulombs, and the plate separation given in millimeters should be converted to meters. Charge, \(Q = 0.020 \times 10^{-6} \,\mathrm{C}\). Meanwhile, Separation, \(d = 0.40 \times 10^{-3} \,\mathrm{m}\)

Apply the capacitance formula

Now apply the capacitance formula \(C = \frac{Q}{V}\), where \(Q\) is the charge on the plates and \(V\) is the potential difference. Substitute the given quantities and solve for capacitance, \(C\). \(C = \frac{0.020 \times 10^{-6}}{240}\)

Calculate the capacitance

Calculate the capacitance by dividing the charge by the potential difference. \(C = \frac{0.020 \times 10^{-6}}{240}\) \(C = 8.333 \times 10^{-11} \,\mathrm{F}\)

Express the result

Finally, express the capacitance to an appropriate number of significant figures and report the result in standard units: Capacitance, \(C = 8.33 \times 10^{-11} \,\mathrm{F}\)