Chapter 23 Flashcards

(15 cards)

1
Q

23-1

A

The electric flux oIo through a surface is the amount of electric field that pierces the surface.

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2
Q

23-1

A

The area vector dA_vector for an area element (patch element) on a surface is a vector that is perpendicular to the element and has a magnitude equal to the area dA of the element.

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3
Q

23-1

A

The electric flux doIo through a patch element with area vector dA_vector is given by a dot product:
doIo = E_vector (dot) dA_vector .

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4
Q

23-1

A

The total flux through a surface is given by oIo = S_integral
E_vector (dot) dA_vector (total flux) where the integration is carried out over the surface.

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5
Q

23-1

A

The net flux through a closed surface (which is used in Gauss’ law) is given by oIo = So_circle_in_integral
E_vector (dot) dA_vector (net flux) where the integration is carried out over the entire surface.

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6
Q

23-2

A

Gauss’ law relates the net flux oIo penetrating a closed surface to the net charge q_enc enclosed by the surface:
E0*oIo = q_enc (Gauss’ law).

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7
Q

23-2

A

Gauss’ law can also be written in terms of the electric field piercing the enclosing Gaussian surface: E0*So_circle_in_integral
E_vector (dot) dA_vector (net flux) = q_enc (Gauss’ law).

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8
Q

23-3

A

An excess charge on an isolated conductor is located entirely on the outer surface of the conductor.

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9
Q

23-3

A

The internal electric field of a charged, isolated conductor is zero, and the external field (at nearby points) is perpendicular to the surface and has a magnitude that depends on the surface charge density o-: E = o- / E0

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10
Q

23-4

A

The electric field at a point near an infinite line of charge (or charged rod) with uniform linear charge density lambda is perpendicular to the line and has magnitude
E = lambda / 2piE0*r (line of charge),
where r is the perpendicular distance from the line to the point.

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11
Q

23-5

A

The electric field due to an infinite nonconducting sheet with uniform surface charge density o- is perpendicular to the plane of the sheet and has magnitude
E = o- / 2*E0 (nonconducting sheet of charge).

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12
Q

23-5

A

he external electric field just outside the surface of an isolated charged conductor with surface charge density o- is perpendicular to the surface and has magnitude
E = o- / E0 (external, charged conductor).
Inside the conductor, the electric field is zero.

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13
Q

23-6

A

Outside a spherical shell of uniform charge q, the electric field due to the shell is radial (inward or outward, depending on the sign of the charge) and has the magnitude
E = (1 / 4piE0) * (q / r^2) (outside spherical shell),
where r is the distance to the point of measurement from the center of the shell. The field is the same as though all of the charge is concentrated as a particle at the center of the shell.

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14
Q

23-6

A

Inside the shell, the field due to the shell is zero.

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15
Q

23-6

A

Inside a sphere with a uniform volume charge density, the
field is radial and has the magnitude
E = (1 / 4piE0) * (q / R^3) * r (inside sphere of charge), where q is the total charge, R is the sphere’s radius, and r is the radial distance from the center of the sphere to the point of measurement.

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