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Chapter 17: Problem 68

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

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

Step by step solution

01

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, \(d = 0.40 \times 10^{-3} \,\mathrm{m}\)
02

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, \(C\). \(C = \frac{0.020 \times 10^{-6}}{240}\)
03

Calculate the capacitance

Calculate the capacitance by dividing the charge by the potential difference. \(C = \frac{0.020 \times 10^{-6}}{240}\) \(C = 8.333 \times 10^{-11} \,\mathrm{F}\)
04

Express the result

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

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Most popular questions from this chapter

The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.
A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
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Draw some electric field lines and a few equipotential surfaces outside a positively charged conducting cylinder. What shape are the equipotential surfaces?
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