By measuring the potential difference between two small probes separated by a distance ΔL, scientists are able to obtain data to plot electric field patterns caused by charges. By rotating the probes, the direction of the electric field can be obtained. From the following two measurements, find the location of the positive charge Q. At the origin, the probe measures a maximum electric field in the direction ux - √3uy. If the probes are moved to the point (0, 1), the maximum electric field is in the direction ux - uy.
Answers
Answer:
Explanation:
In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. (See Figure 1.)
Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either ΔV or E can be used to describe any charge distribution. ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar quantity and has no direction, while E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by E below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from point A to point B.
But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.
The work done by the electric field in Figure 1 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is
W = −ΔPE = −qΔV.
The potential difference between points A and B is
−ΔV = −(VB − VA) = VA − VB = VAB.
Entering this into the expression for work yields W = qVAB.