Question

Calculate the value of electric field intensity at surface of conductor

Answer 1

Answer:

The electric field intensity at surface of conductor is

$\frac{q}{4\pi \r {r}^{2} } e$

Explanation:

According to Gauss's Theorem,

$ln(e \times ds) = ln(e \times nds) = \frac{q}{e.} \\ e \: ln(ds) = \frac{q}{e.} \\ e(4\pi {r}^{2} ) = \frac{q}{e.} \\ e = \frac{q}{4\pi {r}^{2} e.}$

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