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Chapter 17: Problem 61

Two metal spheres have charges of equal magnitude, $3.2 \times 10^{-14} \mathrm{C},$ but opposite sign. If the potential difference between the two spheres is \(4.0 \mathrm{mV},\) what is the capacitance? [Hint: The "plates" are not parallel, but the definition of capacitance holds.]

Short Answer

Expert verified
Answer: The capacitance between the two metal spheres is \(8 \times 10^{-12} \mathrm{F}\).

Step by step solution

01

Write down the definition of capacitance

The definition of capacitance is given by the formula: \(C = \frac{Q}{V}\) where C is the capacitance, Q is the charge, and V is the potential difference.
02

Convert the potential difference to volts

The potential difference between the spheres is given as \(4.0 \mathrm{mV}\). To use it in our calculation, we need to convert it to volts. 1 mV = \(10^{-3}\) V So, \(4.0 \mathrm{mV} = 4.0 \times 10^{-3} \mathrm{V}\)
03

Apply the known values

We are given the charge on the spheres which is \(3.2 \times 10^{-14} \mathrm{C}\) and the potential difference converted to volts which is \(4.0 \times 10^{-3} \mathrm{V}\). Substitute the values into the capacitance formula: \(C = \frac{3.2 \times 10^{-14} \mathrm{C}}{4.0 \times 10^{-3} \mathrm{V}}\)
04

Calculate the capacitance

Now, perform the division to find the value of the capacitance: \(C = \frac{3.2 \times 10^{-14}}{4.0 \times 10^{-3}} = 8 \times 10^{-12} \mathrm{F}\) The capacitance of the two metal spheres is \(8 \times 10^{-12} \mathrm{F}\).

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Most popular questions from this chapter

The bottom of a thundercloud is at a potential of $-1.00 \times 10^{8} \mathrm{V}\( with respect to Earth's surface. If a charge of \)-25.0 \mathrm{C}$ is transferred to the Earth during a lightning strike, find the electric potential energy released. (Assume that the system acts like a capacitor-as charge flows, the potential difference decreases to zero.)
A 2.0 - \(\mu\) F capacitor is connected to a \(9.0-\mathrm{V}\) battery. What is the magnitude of the charge on each plate?
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]
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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