Question

A 30-n/c uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet. the surface charge density in c/m2 on the left and right faces, respectively, are

Answer 1

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Below are the choices that can be found from other sources:

A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2
B. +2.7 × 10−9 C/m2; −2.7 × 10−9 C/m2
C. −5.3 × 10−9 C/m2; +5.3 × 10−9 C/m2
D. +5.3 × 10−9 C/m2; −5.3 × 10−9 C/m2
E. 0; 0

The answer is A. −2.7 × 10−9 C/m2; +2.7 × 10−9 C/m2

Answer 2

Answer:

$\sigma_{left} = -2.7 \times 10^{-10} C/m^2$

$\sigma_{right} = 2.7 \times 10^{-10} C/m^2$

Explanation:

As we know that electric field due to a large sheet is given by the formula

$E = \frac{\sigma}{2\epsilon_0}$

here we know that

$\sigma$ = charge density

now we know that here the field is given for a point between two charged plates

so net electric field is given as

$E = \frac{\sigma}{\epsilon_0}$

$\sigma = E \epsilon_0$

$\sigma = (30)(8.85 \times 10^{-12})$

$\sigma = 2.7 \times 10^{-10} C/m^2$

now here electric field is indicating towards left so here charge density on left side is negative while on right side it must be positive