A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly

Question

A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly distributed along the upper half, and a charge of -q is distributed along the lower half, as shown in Fig. 23-34b. Find the magnitude and direction of the electric field E at P, the center of the semicircle. b) The direction of the electric field at P is: Horizontally to the left Vertically upwards Straight out of the computer screen Straight into the computer screen Horizontally to the right Vertically downwards ????? which one?

Explanation / Answer

By seeing the photo , you can see that the problem has right/left symmetry. So there is no E-field in the middle in the horizontal direction. To get the vertical component, you need some calculus. Divide the circle into little charges dq. The magnitude of the charges is the angular charge density (Q divided by pi radians in a semicircle) times d-theta. That little dq induces an E-field at the center: E = k dq / r^2. But you only want the vertical component, so multiply that times sine theta. Now integrate. When you pull out the constants, you're left integrating sine theta d theta around the semi circle (zero to pi radians) Do that for the two halves. Then add the E-field vectors, making sure to consider whether they are pointing in the same direction or opposite. To get the force, just multiply the E-field by the test charge in the middle.

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