Electrostatics
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Last updated 7:40 PM on 4/16/23
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10 Terms
1
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Negative charges
________ are forced in the direction opposite the field.
2
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Gausss law
________ states that the net electric flux through a closed surface is equal to the charge enclosed divided by ε0.
3
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Positive charges
________ are forced from high to low potential.
4
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Flux
________ only exists if the electric field lines penetrate straight through a surface.
5
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Equipotential Lines
________: Lines that illustrate every point at which a charged particle would experience a given potential.
6
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Electric Field
A property of a region of space that applies a force to charged objects in that region of space
7
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Electric Potential
Potential energy provided by an electric field per unit charge; also called voltage
8
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Equipotential Lines
Lines that illustrate every point at which a charged particle would experience a given potential
9
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A very long cylindrical conductor is surrounded by a very long cylindrical conducting shell. A length L of the inner conductor carries positive charge Q. The same length L of the outer shell carries total charge −3Q.
How much charge is distributed on a length L of the outside surface of the outer shell?
(A) none
(B) −Q
(C) −2Q
(D) −3Q
(E) −4Q
How much charge is distributed on a length L of the outside surface of the outer shell?
(A) none
(B) −Q
(C) −2Q
(D) −3Q
(E) −4Q
C
We have cylindrical symmetry, so use Gauss’s law. Consider a Gaussian surface drawn within the outer shell. Inside a conducting shell, the electric field must be zero, as discussed in the chapter. By Gauss’s law, this means the Gaussian surface we drew must enclose zero net charge. Because the inner cylinder carries charge +Q, the inside surface of the shell must carry charge −Q to get zero net charge enclosed by the Gaussian surface. What happens to the −2Q that is left over on the conducting shell? It goes to the outer surface
We have cylindrical symmetry, so use Gauss’s law. Consider a Gaussian surface drawn within the outer shell. Inside a conducting shell, the electric field must be zero, as discussed in the chapter. By Gauss’s law, this means the Gaussian surface we drew must enclose zero net charge. Because the inner cylinder carries charge +Q, the inside surface of the shell must carry charge −Q to get zero net charge enclosed by the Gaussian surface. What happens to the −2Q that is left over on the conducting shell? It goes to the outer surface
10
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Two identical positive charges Q are separated by a distance a.
What is the electric field at a point halfway between the two charges?
(A) kQ/a2
(B) 2kQ/a2
(C) zero
(D) kQQ/a2
\
What is the electric field at a point halfway between the two charges?
(A) kQ/a2
(B) 2kQ/a2
(C) zero
(D) kQQ/a2
\
C—
Electric field is a vector. Look at the field at the center due to each charge. The field due to the left-hand charge points away from the positive charge; i.e., to the right; the field due to the right-hand charge points to the left. Because the charges are equal and are the same distance from the center point, the fields due to each charge have equal magnitudes. So the electric field vectors cancel! E = 0.
Electric field is a vector. Look at the field at the center due to each charge. The field due to the left-hand charge points away from the positive charge; i.e., to the right; the field due to the right-hand charge points to the left. Because the charges are equal and are the same distance from the center point, the fields due to each charge have equal magnitudes. So the electric field vectors cancel! E = 0.
