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An electron moves away from a proton. Describe how the potential it encounters changes. Describe how its potential energy is changing.

Short Answer

Expert verified
Answer: As the electron moves away from the proton, the electric potential decreases and the electric potential energy of the system also decreases.

Step by step solution

01

Explain the formula for electric potential

The electric potential between two charges is given by the formula: V = k * Q / r where V is the electric potential, k is the electrostatic constant (approximately 8.99 * 10^9 N m^2 C^-2), Q is the charge (in this case, the charge of a proton), and r is the distance between the charges.
02

Describe how the electric potential changes when the electron moves away from the proton

As the distance r between the electron and the proton increases, the electric potential V will decrease according to the formula V = k * Q / r. This means that the electron encounters a lower potential as it moves away from the proton.
03

Explain the formula for electric potential energy

The electric potential energy between two charges is given by the formula: U = k * Q1 * Q2 / r where U is the electric potential energy, Q1 and Q2 are the charges (in this case, the charges of the electron and proton), and r is the distance between the charges.
04

Describe how the electric potential energy changes when the electron moves away from the proton

As the distance r between the electron and the proton increases, their electric potential energy U will decrease according to the formula U = k * Q1 * Q2 / r. This means that the potential energy of the system is lower as the electron moves away from the proton.

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Most popular questions from this chapter

Four identical point charges \((+1.61 \mathrm{nC})\) are placed at the corners of a rectangle, which measures \(3.00 \mathrm{~m}\) by \(5.00 \mathrm{~m}\). If the electric potential is taken to be zero at infinity, what is the potential at the geometric center of this rectangle?

A particle with a charge of \(+5.0 \mu C\) is released from rest at a point on the \(x\) -axis, where \(x=0.10 \mathrm{~m}\). It begins to move as a result of the presence of a \(+9.0-\mu C\) charge that remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point \(x=0.20 \mathrm{~m} ?\)

A classroom Van de Graaff generator accumulates a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) on its spherical conductor, which has a radius of \(10.0 \mathrm{~cm}\) and stands on an insulating column. Neglecting the effects of the generator base or any other objects or fields, find the potential at the surface of the sphere. Assume that the potential is zero at infinity.

How much work would be done by an electric field in moving a proton from a point at a potential of \(+180 . \mathrm{V}\) to a point at a potential of \(-60.0 \mathrm{~V} ?\)

Two fixed point charges are on the \(x\) -axis. A charge of \(-3.00 \mathrm{mC}\) is located at \(x=+2.00 \mathrm{~m}\) and a charge of \(+5.00 \mathrm{mC}\) is located at \(x=-4.00 \mathrm{~m}\) a) Find the electric potential, \(V(x),\) for an arbitrary point on the \(x\) -axis. b) At what position(s) on the \(x\) -axis is \(V(x)=0 ?\) c) Find \(E(x)\) for an arbitrary point on the \(x\) -axis.

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