LIM YUNG KUO PROBLEMS AND SOLUTIONS ON ELECTROMAGNETISM PDF

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Eletromagnetismo 7. If it collapses to a drop of radius 1 mm, what is the change of its electrostatic energy? Solution: Noting that p is a function of only the radius r, we can take a concentric spherical surface of radius r as the Gaussian surface in accordance with the symmetry requirement. The charge dq on this area element will produce at the point P an electric field which is approximately that due to a uniformly charged infinite plate, namely, where n is a unit vector normal to ds in the outward direction.

Hence, if we take E2p as the electric field at P due to all the charges on the spherical surface except the element ds, we must have Therefore, As P is close to ds, E2p may be considered as the field strength at ds due to the charges of the spherical surface. Elceirorioiier 23 a Find the field E at the center of the hollow sphere.

This of course includes the center of the hollow. Con- sider an arbitrary sphere of radius R with a uniform charge density p. Using 1 and the superposition theorem, we obtain Now consider the problem in hand. Determine whether 4 or some derivative of it is discontinuous across the layer and find the discontinuity.

On this surface there is a uniform dipole layer T and a uniform surface charge density u. Find T and u so that the poten- tial inside the surface will be just that of the charge q , while the potential outside will be zero. You may make use of whatever you know about the potential of a surface charge. We choose cylindrical coordinates R, 8 , z such 25 that P is on the z-axis. Thus If the potential at infinity is zero, then the potential outside the spherical surface will be zero everywhere.

One terminal is connected to ground and the other terminal is insulated and not connected to anything. What happens to the potential on the free plate when the plates are separated?

Express the potential VZ in terms of the potential V1. Solution: In the process of separation the charge on the insulated plate is kept constant. The area of each plate is A. If the voltage difference between the outside plates is kept constant a;? From Fig.

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Problems and Solutions on Electromagnetism - Lim Yung Kuo

Eletromagnetismo 7. If it collapses to a drop of radius 1 mm, what is the change of its electrostatic energy? Solution: Noting that p is a function of only the radius r, we can take a concentric spherical surface of radius r as the Gaussian surface in accordance with the symmetry requirement. The charge dq on this area element will produce at the point P an electric field which is approximately that due to a uniformly charged infinite plate, namely, where n is a unit vector normal to ds in the outward direction. Hence, if we take E2p as the electric field at P due to all the charges on the spherical surface except the element ds, we must have Therefore, As P is close to ds, E2p may be considered as the field strength at ds due to the charges of the spherical surface. Elceirorioiier 23 a Find the field E at the center of the hollow sphere. This of course includes the center of the hollow.

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