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jok3333 [9.3K]
3 years ago
11

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

opposite signs. The amount of charge per unit area is given by |σ| = 6.8 ✕ 10-22 C/m2, with the negatively charged plate on the left. What are the magnitude and direction of the electric field vector E in unit-vector notation at the following locations?
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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