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Two points, \(A\) and \(B\), are separated by \(0.016 \mathrm{~m}\). The potential at \(A\) is \(+95 \mathrm{~V}\), and that at \(B\) is \(+28 \mathrm{~V}\). Find the magnitude and direction of the constant electric field between the points.

Short Answer

Expert verified
The electric field is approximately \(4187.5 \mathrm{~V/m}\) directed from \(A\) to \(B\).

Step by step solution

01

Understanding the Problem

We are given two points, \(A\) and \(B\), with known potentials, \(V_A = +95 \mathrm{~V}\) and \(V_B = +28 \mathrm{~V}\), separated by a distance of \(0.016 \mathrm{~m}\). We need to find the magnitude and direction of the electric field between them.
02

Recall the Relationship Between Electric Field and Potential Difference

The electric field \(E\) between two points in a uniform electric field is given by the equation: \[ E = -\frac{\Delta V}{d} \]where \(\Delta V = V_B - V_A\) is the potential difference between the points, and \(d\) is the distance between the points.
03

Calculate the Potential Difference

Calculate the potential difference, \(\Delta V\):\[ \Delta V = V_B - V_A = 28 \mathrm{~V} - 95 \mathrm{~V} = -67 \mathrm{~V} \]
04

Calculate the Magnitude of the Electric Field

Substitute the potential difference and distance into the formula to find the electric field:\[ E = -\frac{-67 \mathrm{~V}}{0.016 \mathrm{~m}} = \frac{67 \mathrm{~V}}{0.016 \mathrm{~m}} \approx 4187.5 \mathrm{~V/m} \]The negative signs cancel out, resulting in a positive electric field magnitude.
05

Determine the Direction of the Electric Field

The direction of the electric field is from the point with higher potential to the point with lower potential. Since \(V_A > V_B\), the electric field is directed from point \(A\) to point \(B\).

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Key Concepts

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

Potential Difference
The concept of potential difference is central to understanding electric fields. It refers to the difference in electric potential between two points in space. In simpler terms, it's the work needed to move a unit charge from one point to another. Electric potential is measured in volts (V), and when we talk about potential difference, we're usually discussing two specific points, such as points A and B.
The potential difference, \(\Delta V\), between two points A and B is given by the equation:
  • \(\Delta V = V_B - V_A\)
In this problem, the potential at point A is 95 volts, and at point B, it's 28 volts. So, the potential difference is calculated as \(\Delta V = 28 \mathrm{~V} - 95 \mathrm{~V} = -67 \mathrm{~V}\).Understanding this negative sign is crucial. It indicates that as we move from A to B, the electric potential decreases by 67 volts.
The negative potential means point B is at a lower potential than point A, which directly affects the direction of the electric field.
Electric Potential
Electric potential, often simply called voltage, represents the electric potential energy per unit charge at a point in space. Think of it like an "electric height" from which charges "fall," with higher voltages being like mountain peaks and lower voltages like valleys.
  • A charge at a point with higher potential has more stored energy.
  • The unit of electric potential is the volt (V).
In this exercise, points A and B have different potentials, with A having a higher potential (95 V) than B (28 V). This difference creates potential energy for charges moving between these points. Charges will naturally move from higher potential (A) to lower potential (B), much like how water flows downhill.
The electric potential is a scalar quantity, which means it doesn't have direction. But, potential differences between points do result in directional effects, such as forces and fields acting on charges within that field.
Uniform Electric Field
A uniform electric field is a field in which the electric force exerted on a charge is the same no matter where the charge is located in the field. This means the field lines are parallel and equally spaced.
Electric fields are typically represented by the symbol \(E\), with units of volts per meter (V/m). In a uniform field, the electric potential difference between two points is directly related to the field strength and the distance between them:
  • \(E = -\frac{\Delta V}{d}\)
In our problem, the field is uniform between points A and B. Using the potential difference \(\Delta V = -67\) V and the distance \(d = 0.016\) meters, the magnitude of the electric field is calculated as \[E = -\frac{-67 \mathrm{~V}}{0.016 \mathrm{~m}} = 4187.5 \mathrm{~V/m}\]The uniform electric field indicates that no matter where you place a charge between A and B, it will experience the same electric force. The direction, described by the negative potential difference, is from A (higher potential) to B (lower potential), showing how charges respond naturally to electric forces.

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Most popular questions from this chapter

A positive charge \(+q_{1}\) is located to the left of a negative charge \(-q_{2} .\) On a line passing through the two charges, there are two places where the total potential is zero. The first place is between the charges and is \(4.00 \mathrm{~cm}\) to the left of the negative charge. The second place is \(7.00 \mathrm{~cm}\) to the right of the negative charge. (a) What is the distance between the charges? (b) Find \(q_{1} / q_{2}\), the ratio of the magnitudes of the charges.

Two charges \(A\) and \(B\) are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge \(\mathrm{A}\) is \(0.18 \mathrm{~m}\) from the spot, while charge \(\mathrm{B}\) is \(0.43 \mathrm{~m}\) from it. Find the ratio \(q_{\mathrm{B}} / q_{\mathrm{A}}\) of the charges.

A particle is uncharged and is thrown vertically upward from ground level with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). As a result, it attains a maximum height \(h\). The particle is then given a positive charge \(+q\) and reaches the same maximum height \(h\) when thrown vertically upward with a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The electric potential at the height \(h\) exceeds the electric potential at ground level. Finally, the particle is given a negative charge \(-q\). Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height \(h\). In all three situations, be sure to include the effect of gravity.

Suppose that the electric potential outside a living cell is higher than that inside the cell by \(0.070 \mathrm{~V}\). How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

Refer to Interactive Solution \(\underline{19.33}\) at to review a method by which this problem can be solved. The electric field has a constant value of \(4.0 \times 10^{3} \mathrm{~V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is \(155 \mathrm{~V}\). Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{~m}\) directly above \(P\), (b) \(3.0 \times 10^{-3} \mathrm{~m}\) directly below \(P,\) (c) \(8.0 \times 10^{-3} \mathrm{~m}\) directly to the right of \(P\).

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