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Introductory Physics Volume Two

Introductory Physics Volume Two

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32 Electric Field 1.8<br />

With the assumptions above, symmetry tells us that the electric field<br />

will point in a direction perpendicular to the surface; for a positive<br />

charge it will point away from the plate:<br />

E<br />

A<br />

σ<br />

E<br />

In the figure a Gaussian surface is indicated. Take it to be a small<br />

cylinder with cross-sectional area A. The electric flux in non-zero only<br />

through the endcaps:<br />

∮<br />

∫<br />

∫<br />

∫<br />

⃗E · dA ⃗ = E · dA ⃗ + E · dA ⃗ + E · dA<br />

⃗<br />

upper cap<br />

sides<br />

= EA + 0 + EA = 2EA<br />

Using Gauss’s law:<br />

2EA = σA<br />

ɛ 0<br />

−→ E =<br />

σ<br />

2ɛ 0<br />

.<br />

lower cap<br />

§ 1.8 Homework<br />

⊲ Problem 1.13<br />

y<br />

7.0 µC<br />

x<br />

0.50 m<br />

0.50 m<br />

2.0 µC<br />

0.50 m<br />

−4.0 µC<br />

Three point charges are located at the corners of an equilateral triangle<br />

as shown. Calculate the net electric force on the 7.0µC charge.

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