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Chapter 17: Problem 90

If an electron moves from one point at a potential of \(-100.0 \mathrm{V}\) to another point at a potential of \(+100.0 \mathrm{V}\) how much work is done by the electric field?

Short Answer

Expert verified
Answer: The work done by the electric field on the electron is \(-3.2 \times 10^{-17} \, \mathrm{J}\).

Step by step solution

01

Write down the given information

We are given the initial electric potential, \(V_i = -100.0 \mathrm{V}\), and the final electric potential, \(V_f = +100.0 \mathrm{V}\). We need to find out the work done by the electric field (\(W\)).
02

Write down the formula to calculate the work done by the electric field

According to the work-energy theorem, the work done by the electric field is equal to the change in the electron's potential energy. The formula for the work done by the electric field is: $$W = q \Delta V$$ where \(q\) is the charge of the electron and \(\Delta V\) is the change in electric potential.
03

Calculate the change in electric potential

The change in electric potential is the final electric potential minus the initial electric potential: $$\Delta V = V_f - V_i$$ Substitute the given values into the equation: $$\Delta V = (+100.0 \mathrm{V}) - (-100.0 \mathrm{V}) = 200.0 \mathrm{V}$$
04

Identify the charge of the electron

The charge of the electron, \(q\), is a known constant value: $$q = -1.6 \times 10^{-19} \, \mathrm{C}$$ Note that the charge is negative, as electrons are negatively charged particles.
05

Calculate the work done by the electric field

Using the formula from Step 2, substitute the values for \(q\) and \(\Delta V\) to find the work done by the electric field: $$W = q \Delta V$$ $$W = (-1.6 \times 10^{-19} \, \mathrm{C}) (200.0 \mathrm{V})$$ Calculate the work: $$W = -3.2 \times 10^{-17} \, \mathrm{J}$$ So, the work done by the electric field on the electron is \(-3.2 \times 10^{-17} \, \mathrm{J}\).

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Most popular questions from this chapter

An alpha particle (helium nucleus, charge \(+2 e\) ) starts from rest and travels a distance of \(1.0 \mathrm{cm}\) under the influence of a uniform electric field of magnitude \(10.0 \mathrm{kV} / \mathrm{m}\) What is the final kinetic energy of the alpha particle?
Two parallel plates are \(4.0 \mathrm{cm}\) apart. The bottom plate is charged positively and the top plate is charged negatively, producing a uniform electric field of \(5.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) in the region between the plates. What is the time required for an electron, which starts at rest at the upper plate, to reach the lower plate? (Assume a vacuum exists between the plates.)
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.
Figure 17.31 b shows a thundercloud before a lightning strike has occurred. The bottom of the thundercloud and the Earth's surface might be modeled as a charged parallel plate capacitor. The base of the cloud, which is roughly parallel to the Earth's surface, serves as the negative plate and the region of Earth's surface under the cloud serves as the positive plate. The separation between the cloud base and the Earth's surface is small compared to the length of the cloud. (a) Find the capacitance for a thundercloud of base dimensions \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\) located \(550 \mathrm{m}\) above the Earth's surface. (b) Find the energy stored in this capacitor if the charge magnitude is \(18 \mathrm{C}\).
A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
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