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
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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:
\(\sigma_{left} = -2.7 \times 10^{-10} C/m^2\)
\(\sigma_{right} = 2.7 \times 10^{-10} C/m^2\)
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