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Chapter 23: Problem 23

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{~m}\) ) and point \(b\) (at \(x=0.90 \mathrm{~m}\) ) is \(240 \mathrm{~V}\). (a) Which point, \(a\) or \(b,\) is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

Short Answer

Expert verified
(a) Point \(a\) is at a higher potential. (b) The magnitude of electric field \(E\) is \(800 N/C\). (c) The work done on the point charge by the electric field is \(-0.048 J\).

Step by step solution

01

Identify the Higher Potential Point

The electric field is directed from the higher potential towards the lower potential in the negative x-direction. Thus, point \(a\) is at a higher potential than point \(b\).
02

Calculate the Magnitude of the Electric Field

The electric field \(E\) can be found using the equation \(E = \Delta V / \Delta x\), where \(\Delta V\) is the difference in electric potential and \(\Delta x\) is the separation between the points. Here, \(\Delta x = 0.30 m\) (0.90m - 0.60m) and \(\Delta V = 240 V\). Substituting these into the equation gives \(E = 240V / 0.30m = 800 N/C\).
03

Calculate the Work Done on the Charge

Work done can be calculated using the equation \(W = q\Delta V\), where \(q\) is the charge and \(\Delta V\) is the potential difference. Here, \(q = -0.200 \mu C = -0.200 \times 10^{-6} C\) and \(\Delta V = 240 V\). Substituting these into the equation gives \(W = -0.200 \times 10^{-6} C \times 240V = -0.048 J\). Since the work done is negative, this indicates that work is done by the field to move the charge from point \(b\) to \(a\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Electric Field
Imagine a grid of straight, parallel lines, each pointing in the same direction and evenly spaced. This is a visual representation of a uniform electric field. Such a field exerts the same magnitude and direction of force on charges throughout its extent.

When you place a positive test charge within this grid, it feels an identical push or pull no matter where it is located in the field. Understanding uniform electric fields is crucial in predicting the behavior of charges in various situations, including the basic operation of electronic devices, where components are often subject to constant electric fields.
Electric Potential
The electric potential at a point in space, often measured in volts (V), is akin to the height of a hill in a gravitational field. It represents the potential energy per unit charge. If a charge were a ball, higher potential could be thought of as the top of the hill; rolling down, the ball would gain speed, just as a charge would accelerate moving towards lower potential.

A crucial aspect for students to grasp is that electric potential allows us to compute the work done by or on charges as they move through an electric field without directly calculating the forces involved.
Work Done by Electric Field
The concept of work in physics is quite different from its everyday use. It is a measure of energy transfer. For a charge in an electric field, work is done by the field when the charge moves, analogous to the work done by gravity on a rolling ball.

Work and Potential Difference

In the context of electric fields, the work done by the field on a charge is directly related to the potential difference the charge moves across. This is an invaluable tool when analyzing circuits or understanding how energy is transferred in electric fields.
Magnitude of Electric Field
The magnitude of an electric field, expressed in newtons per coulomb (N/C), tells us the force experienced by a positive unit charge placed in the field. It is a vector quantity, meaning it has both magnitude and direction, and in a uniform electric field, this magnitude is a constant value.

The magnitude and direction of an electric field can also determine how electrons move in a wire, how particles get accelerated in a particle accelerator, or help in medical applications such as the designing of defibrillators. Hence, calculating the magnitude correctly, as demonstrated in our original example, is fundamental for applications ranging from electronics to particle physics.

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Most popular questions from this chapter

CP Deflecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying cqual but opposite charges. Typical dimensions are about \(3.0 \mathrm{~cm}\) on a side, with a separation of about \(5.0 \mathrm{~mm} .\) The potential difference between the plates is \(25.0 \mathrm{~V}\). The plates are close enough that we can ignore fringing at the cnds. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Two point charges \(q_{1}=\) \(+2.40 \mathrm{nC}\) and \(q_{2}=-6.50 \mathrm{nC}\) are \(0.100 \mathrm{~m}\) apart. Point \(A\) is midway between them; point \(B\) is \(0.080 \mathrm{~m}\) from \(q_{1}\) and \(0.060 \mathrm{~m}\) from \(q_{2}\) (Fig. E.23.19). Take the clectric potential to be zero at infinity. Find (a) the potential at point \(A ;\) (b) the potential at point \(B ;(\mathrm{c})\) the work done by the electric field on a charge of \(2.50 \mathrm{nC}\) that travels from point \(B\) to point \(A\).

A long metal cylinder with radius \(a\) is supported on an insulating stand on the axis of a long. hollow, metal tube with radius \(b\). The positive charge per unit length on the inner cylinder is \(\lambda\), and there is an equal negative charge per unit length on the outer cylinder. (a) Calculate the potential \(V(r)\) for (i) \(rb .\) (Hint: The net potcntial is the sum of the potentials due to the individual conductors.) Take \(V=0\) at \(r=b .\) (b) Show that the potential of the inner cylinder with respect to the outer is \(V_{a b}=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{b}{a}\) (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the cylinders has magnitude E(r)=\frac{V_{a b}}{\ln (b / a)} \frac{1}{r} (d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y^{-},\) and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?
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