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Chapter 21: Q1CP (page 870)
In Figure 21.15 the magnitude of the electric field is , and the field is at an angle of to the outward-going normal. What is the flux on the small rectangle whose dimensions are by ?
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Along the path shown in Figure 21.46 the magnetic field is measured and is found to be uniform in magnitude and always tangent to the circular path. If the radius of the path is and along the path is , what are magnitude and direction of the current enclosed by the path?
The electric field has been measured to be horizontal and to the right everywhere on the closed box as shown in Figure 21.66. All over the left side of box and all over the right (slanting) side of box .On the top the average field is , on the front and back the average field is and on the bottom the average field is .How much charge is inside the box? Explain briefly.
A lead nucleus is spherical with a radius of about . The nucleus contains protons (and typically neutrons). Because of their motions the protons can be considered on average to be uniformly distributed throughout the nucleus. Base on the net flux at the surface of the nucleus, calculate the divergence of the electric field as electric flux per unit volume. Repeat the calculation at a radius of . (You can use Gauss鈥檚 law to determine the magnitude of the electric field at this radius.) Also calculate the quantity inside the nucleus.
The electric field is horizontal and has the values indicated on the surface of cylinder as shown in Figure 21.65. What can you deduce from this pattern of electric field? Include a numerical result.
Figure 21.62 shows a box on who surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of electric field is 400 V/m and on the right face the magnitude of electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux and total amount of charge that is inside the box.
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