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Chapter 23: Problem 24

A small sphere with charge \(q=-5.00 \mu \mathrm{C}\) is moving in a uniform electric field that has no \(y\) - or \(z\) -component. The only force on the sphere is the force exerted by the electric field. Point \(A\) is on the \(x\) -axis at \(x=-0.400 \mathrm{~m},\) and point \(B\) is at the origin. At point \(A\) the sphere has kinetic energy \(K_{A}=8.00 \times 10^{-4} \mathrm{~J},\) and at point \(B\) its kinetic energy is \(K_{B}=3.00 \times 10^{-4} \mathrm{~J}\). (a) What is the potential difference \(V_{A B}=V_{A}-V_{B} ?\) Which point, \(A\) or \(B,\) is at higher potential? (b) What are the magnitude and direction of the electric field?

Short Answer

Expert verified
The potential difference \(V_{A B}=V_{A}-V_{B}\) = -100.0 V, with point B at a higher potential than point A. The magnitude of the electric field is 250 N/C and it points from B to A.

Step by step solution

01

Calculate the potential difference (VA-VB) and Identify the Point at Higher Potential

The work done by the electric field on the charge \( q \) is equal to the change in the kinetic energy of the charge:\[W = K_B - K_A = q(V_B - V_A)\]We can rearrange this equation to solve for \( V_B - V_A \):\[V_B - V_A = \frac{K_B - K_A}{q}\]Substitute the given values into the equation:\[V_B - V_A = \frac{3.00 \times 10^{-4} J - 8.00 \times 10^{-4} J}{-5.00 \times 10^{-6} C} \]Resulting in: \(V_B - V_A = 100.0 V\). Since \( V_B - V_A > 0 \), point B is at a higher potential.
02

Calculate the Magnitude and Direction of the Electric Field

We know that the magnitude of the electric field is the potential difference divided by the distance. So we can write:\[E= \frac{|V_B - V_A|}{|x_B - x_A|}\]Substitute the values into the equation:\[E = \frac{100.0 V}{0.400 m} = 250 N/C\]Now the direction of the electric field is from point of high potential to low potential. Hence, the electric field points from B to A.

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Key Concepts

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

Electric Potential
Electric potential is a fundamental concept in understanding electric fields and the forces they exert. In simple terms, electric potential represents the potential energy per unit charge at a specific point in space due to an electric field. It is analogous to gravitational potential energy in a gravitational field but applies to electric charges instead. The value of electric potential at a point is measured in volts (V).

When a charge moves within an electric field, its potential energy changes. This change in energy is related to the electric potential difference between the starting and ending points.
  • Positive charges naturally move from areas of high potential to low potential.
  • Negative charges, conversely, move from low to high potential.
In the original exercise, the potential difference between points A and B is calculated to be 100V, with point B at a higher potential than point A. Understanding electric potential helps explain how charges move and the work done on them by the electric field.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving possesses kinetic energy, which depends on its mass and speed. The faster an object moves, or the more massive it is, the more kinetic energy it holds. The formula for calculating kinetic energy is given by: \[ KE = \frac{1}{2}mv^2 \]- Where \( KE \) represents the kinetic energy, \( m \) is the mass, and \( v \) is the velocity of the object.

In the context of the electric field exercise, we see the kinetic energy of a charged particle changing as it moves between points A and B. At point A, the kinetic energy is higher than at point B. The reduction in kinetic energy indicates that the electric field does work on the charge, which is associated with a change in the electric potential energy as the charge moves through the field. The change in kinetic energy gives us insight into how the electric field interacts with charged particles.
Potential Difference
The potential difference, also known as voltage, is the difference in electric potential between two points in an electric field. It tells us how much work is done by the electric field in moving a charge from one point to another. The potential difference is crucial for understanding how charges are influenced by electric fields.

Mathematically, the potential difference between two points A and B can be expressed as: \[ V_{AB} = V_A - V_B \]In the given problem, solving for this difference gives us an understanding of the field's influence on the moving charge. Point B is found to be at a higher potential, meaning work is done by the field when a negative charge moves towards a region of higher potential.

Potential difference is often harnessed in electrical circuits, where it drives the flow of electrons through conductors, powering electronic devices.
  • A positive potential difference implies work done on the charge by the external field.
  • A negative potential difference implies the charge does work against the field.
Understanding potential difference is essential for grasping how electric fields transfer energy and exert forces.

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Most popular questions from this chapter

BIO Electrical Sensitivity of Sharks. Certain sharks can detect an electric field as weak as \(1.0 \mu \mathrm{V} / \mathrm{m}\). To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 V AA battery across these plates, how far apart would the plates have to be?

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil was sprayed in very fine drops (about \(10^{-4} \mathrm{~mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d\). A potential difference \(V_{A B}\) was maintained between the plates, causing a downward electric field between them. Some of the oil drops acquired a negative charge because of frictional effects or because of ionization of the surrounding air by x rays or radioactivity. The drops were observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates remained at rest if the magnitude of its charge was \(q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}\) where \(\rho\) is oil's density. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, Millikan determined the charge on that drop. provided its radius \(r\) was known. (b) Millikan's oil drops were much too small to measure their radii directly. Instead, Millikan determincd \(r\) by cutting off the electric field and measuring the terminal speed \(u_{\mathrm{i}}\) of the drop as it fell. (We discussed terminal speed in Section \(\left.5.3 .\right)\) The viscous force \(F\) on a sphere of radius \(r\) moving at speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When a drop fell at \(v_{1}\), the viscous force just balanced the drop's weight \(w=m g\). Show that the magnitude of the charge on the drop was \(q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{1}^{3}}{2 \rho g}}\) (c) You repeat the Millikan oil-drop experiment. Four of your measured values of \(V_{A B}\) and \(v_{1}\) are listed in the table: $$ \begin{array}{lcccc} \text { Drop } & 1 & 2 & 3 & 4 \\ \hline V_{A B}(\mathrm{~V}) & 9.16 & 4.57 & 12.32 & 6.28 \\ v_{1}\left(10^{-5} \mathrm{~m} / \mathrm{s}\right) & 2.54 & 0.767 & 4.39 & 1.52 \end{array} $$ In your apparatus, the separation \(d\) between the horizontal plates is \(1.00 \mathrm{~mm}\). The density of the oil you use is \(824 \mathrm{~kg} / \mathrm{m}^{3}\). For the viscosity \(\eta\) of air, use the value \(1.81 \times 10^{-5} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the charge \(q\) of each drop. (d) If electric charge is quantized (that is, exists in multiples of the magnitude of the charge of an electron), then the charge on each drop is -ne, where \(n\) is the number of excess electrons on each drop. (All four drops in your table have negative charge.) Drop 2 has the smallest magnitude of charge observed in the experiment, for all 300 drops on which measurements were made, so assume that its charge is due to an excess charge of one electron. Determine the number of excess electrons \(n\) for each of the other three drops. (e) Use \(q=-n e\) to calculate \(e\) from the data for each of the four drops, and average these four values to get your best experimental value of \(e .\)

Two protons, starting several meters apart, are aimed directly at each other with speeds of \(2.00 \times 10^{5} \mathrm{~m} / \mathrm{s},\) measured relative to the earth. Find the maximum electric force that these protons will exert on each other.

A small particle has charge \(-5.00 \mu\) C and mass \(2.00 \times 10^{-4} \mathrm{~kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{~V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{~V}\) The electric force is the only force acting on the particle. The particle has speed \(5.00 \mathrm{~m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

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?
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