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A charge of 8 nC is placed uniformly on a square sheet of nonconducting material of side 22 cm in the yz plane.(d) What is the magnitude of the electric field to the right (x > 0) of the right face of the slab?

Verified Answer :

metchelle

This is a problem dealing with the surface charge density. Based on the given problem above, your charge is 8nc on a square plate of 22 cm x22 cm given that E= (surface charge density)/2ε₀ so answer would be 8nc/22/22/2ε₀. Therefore, the magnitude of the electric field to the right (x>0) of the right face of the slab would be 8nc/22/22/2ε₀.

Answer:

E = 9338.4 N/C

Explanation:

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

\(E = \frac{\sigma}{2\epsilon_0}\)

here we know that

\(\sigma\) = charge density

\(\sigma = \frac{Q}{A}\)

\(\sigma = \frac{Q}{L^2}\)

\(\sigma = \frac{8\times 10^{-9}}{0.22^2}\)

\(\sigma = 1.65 \times 10^{-7}C/m^2\)

now we have

\(E = \frac{1.65 \times 10^{-7}}{2(8.85 \times 10^{-12}}\)

\(E = 9338.4 N/C\)

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