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Chapter 23: Problem 1

A positive charge is released and moves along an electric field line. This charge moves to a position of a) lower potential and lower potential energy. b) lower potential and higher potential energy. c) higher potential and lower potential energy. d) higher potential and higher potential energy.

Short Answer

Expert verified
a) lower potential and lower potential energy b) lower potential and higher potential energy c) higher potential and higher potential energy d) higher potential and lower potential energy Answer: a) lower potential and lower potential energy

Step by step solution

01

Understanding Electric Potential and Potential Energy

Electric potential is the electric potential energy per unit charge. If a charge moves in the direction of the electric field, the potential decreases. On the other hand, if it moves against the electric field, the potential increases. Electric potential energy is the energy stored in a system as a result of the electrostatic interactions between various charges in the system. The electric potential energy of a charge 'q' at a point in space with potential 'V' is given by U=qV. If the potential decreases, the potential energy also decreases for a positive charge. And, if the potential increases, the potential energy increases for a positive charge.
02

Option (a) Analysis

In this option, both potential and potential energy decrease. Since the charge is positive and moves along the electric field, it would cause the potential to decrease. Also, as the potential decreases, the potential energy of a positive charge would also decrease. So, this option is valid.
03

Option (b) Analysis

In this option, potential decreases, but potential energy increases. This is a contradiction, because if potential decreases for a positive charge, its potential energy should also decrease. So, this option is not valid.
04

Option (c) Analysis

In this option, both potential and potential energy increase. This means that a positive charge would have to move against the electric field, which is not the case stated in the problem (the charge moves along the electric field line). So this option is not valid.
05

Option (d) Analysis

In this option, potential increases, but potential energy decreases. This is a contradiction because if potential increases for a positive charge, its potential energy should also increase. So, this option is not valid. The only valid option is: a) lower potential and lower potential energy.

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Most popular questions from this chapter

A total charge of \(Q=4.2 \cdot 10^{-6} \mathrm{C}\) is placed on a conducting sphere (sphere 1 ) of radius \(R=0.40 \mathrm{~m}\). a) What is the electric potential, \(V_{1},\) at the surface of sphere 1 assuming that the potential infinitely far away from it is zero? (Hint: What is the change in potential if a charge is brought from infinitely far away, where \(V(\infty)=0,\) to the surface of the sphere?) b) A second conducting sphere (sphere 2) of radius \(r=0.10 \mathrm{~m}\) with an initial net charge of zero \((q=0)\) is connected to sphere 1 using a long thin metal wire. How much charge flows from sphere 1 to sphere 2 to bring them into equilibrium? What are the electric fields at the surfaces of the two spheres?

A charge \(Q=\) \(+5.60 \mu C\) is uniformly distributed on a thin cylindrical plastic shell. The radius, \(R\), of the shell is \(4.50 \mathrm{~cm}\). Calculate the electric potential at the origin of the \(x y\) -coordinate system shown in the figure. Assume that the electric potential is zero at points infinitely far away from the origin.

A proton is placed midway between points \(A\) and \(B\). The potential at point \(A\) is \(-20 \mathrm{~V}\), and the potential at point \(B\) \(+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

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.

An electric field is established in a nonuniform rod. A voltmeter is used to measure the potential difference between the left end of the rod and a point a distance \(x\) from the left end. The process is repeated, and it is found that the data are described by the relationship \(\Delta V=270 x^{2},\) where \(\Delta V\) has the units \(\mathrm{V} / \mathrm{m}^{2}\). What is the \(x\) -component of the electric field at a point \(13 \mathrm{~cm}\) from the left end?
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