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Chapter 23: Problem 21

Points \(A\) and \(B\) lie within a region of space where there is a uniform electric field that has no \(x\) - or \(z\) -component; only the \(y\) -component \(E_{y}\) is nonzero. Point \(A\) is at \(y=8.00 \mathrm{~cm}\) and point \(B\) is at \(y=15.0 \mathrm{~cm} .\) The potential difference between \(B\) and \(A\) is \(V_{B}-V_{A}=+12.0 \mathrm{~V},\) so point \(B\) is at higher potential than point \(A\). (a) Is \(E_{y}\) positive or negative? (b) What is the magnitude of the electric field? (c) Point \(C\) has coordinates \(x=5.00 \mathrm{~cm}, y=5.00 \mathrm{~cm} .\) What is the potential difference between points \(B\) and \(C ?\)

Short Answer

Expert verified
a) The \(y\) -component of the electric field \(E_{y}\) is positive.\nb) The magnitude of the electric field is \(171 \, V/m.\)\nc) The potential difference between points \(B\) and \(C\) is \(-17.1 \, V.\)

Step by step solution

01

Find the Direction of the Electric Field

A positive potential difference means moving in the direction of the electric field. Since \(B\) is at a higher potential than \(A\), and \(B\) is greater than \(A\) in the \(y\) -direction, this means that the electric field \(E_{y}\) is positive.
02

Find the Magnitude of the Electric Field

The magnitude of the electric field \(E_{y}\) can be found using the equation \( \Delta V = -E\Delta y \), where \( \Delta V = V_B - V_A \) and \( \Delta y = y_B - y_A \). Solving this equation gives \( E_y= -\Delta V/\Delta y = -(V_B - V_A)/(y_B - y_A) = -(+12.0 V)/(15.0 cm - 8.00 cm) = -1.71 V/cm = -171 V/m.\) However, since we found that \(E_{y}\) is positive, the magnitude of the electric field is \(171 \, V/m.\)
03

Calculate the Potential Difference between Points \(B\) and \(C\)

Finally, the potential difference \( \Delta V_{BC} = V_B -V_C \) can be found using the equation \(\Delta V = -E \Delta y\), where \(\Delta y = y_B - y_C\). Substituting the known values gives \(\Delta V_{BC} = -(E_{y})(y_B - y_C) = -171 V/m (15.0 cm - 5.00 cm) = -1.71 V/cm (10.0 cm) = -17.1 V.\) Therefore, the potential difference between points \(B\) and \(C\) is \(-17.1 \, V.\)

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Key Concepts

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

Potential Difference
The potential difference, often referred to as voltage, is a fundamental concept in the field of electricity. It tells us about the energy difference per unit charge between two points in an electric field. Think of it as the electrical 'pressure' that motivates electrons to move from one point to another.

A positive potential difference, as mentioned in the given problem, indicates that one point (point B in this case) is at a higher potential energy than another point (point A). This acts as a motivating 'force' for electrons to move towards the lower potential point when a conductive path is provided. In a battery, for example, the potential difference between the terminals drives the current through a circuit.

One way to visualize potential difference is by thinking about a hill. If point B is at the top of the hill and point A is at the bottom, the ball (representing a charge) will naturally roll down from higher to lower elevation, which is similar to how electrons move from higher to lower electrical potential.
Uniform Electric Field
A uniform electric field is characterized by electric field vectors that have the same magnitude and direction at every point within the field. This makes the field predictable and easy to calculate, much like the effect of gravity near the Earth's surface is considered uniform for practical purposes.

In the context of the exercise, the electric field only has a component along the y-axis, meaning it is uniform in the vertical direction with respect to the problem's coordinate system. A uniform electric field is typically created between two parallel plates with a voltage across them, such as in a parallel-plate capacitor. If you were to place test charges in a uniform electric field, each charge would experience the same force regardless of its position in the field.

Mathematically, the uniformity of the field provides us with the convenience of being able to exclude the shape of the path taken by a charge between two points when calculating work done, making potential difference calculations more straightforward.
Magnitude of Electric Field
The magnitude of an electric field represents the force experienced by a positive test charge placed within the field, per unit of charge. It's measured in volts per meter (V/m) or, equivalently, in newtons per coulomb (N/C).

The magnitude of the electric field within a region tells us how strong or intense the field is. Higher magnitudes mean a greater force on charges in the field. In the given exercise, determining the magnitude of the electric field was crucial to understand how much force a charge would experience when moving from one point to another.

To calculate this magnitude, as shown in the solution steps, we used the potential difference between points and the distance between them. This provides a simple yet powerful way to determine the field's strength. In practical terms, knowing the electric field's magnitude allows engineers to design electrical equipment that can withstand the forces within the field, like the insulation on a power line.

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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?

A total clectric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\) (b) \(24.0 \mathrm{~cm}\) (c) \(12.0 \mathrm{~cm}\)

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) \(-3.0 \mathrm{MV}\) (b) \(+3.0 \mathrm{MV} ;(\mathrm{c})+1.5 \mathrm{MV} ;(\mathrm{d})+1.0 \mathrm{MV}\)

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

An annulus with an inner radius of \(a\) and an outer radius of \(b\) has charge density \(\sigma\) and lies in the \(x y\) -plane with its center at the origin, as shown in Fig. \(\mathbf{P} 2 \mathbf{3 . 8 0}\). (a) Using the convention that the potential vanishes at infinity, determine the potential at all points on the \(z\) -axis. (b) Determine the electric field at all points on the \(z\) -axis by differentiating the potential. (c) Show that in the limit \(a \rightarrow 0, b \rightarrow \infty\) the electric field reproduces the result obtained in Example 22.7 for an infinite plane sheet of charge. (d) If \(a=5.00 \mathrm{~cm}, b=10.0 \mathrm{~cm}\) and the total charge on the annulus is \(1.00 \mu \mathrm{C}\), what is the potential at the origin? (e) If a particle with mass \(1.00 \mathrm{~g}\) (much less than the mass of the annulus) and charge \(1.00 \mu \mathrm{C}\) is placed at the origin and given the slightest nudge, it will be projected along the \(z\) -axis. In this case, what will be its ultimate speed?
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