Question

Due to a conducting infinite plane sheet
having surface charge density = 8.85 uC/mi
at 10 cm electric field intensity in N/C​

Answer 1

Given:

A conducting infinite plane sheet having surface charge density = 8.85 uC/m².

To find:

Field intensity at 10 cm distance?

Calculation:

Applying Gauss' Law:

$\displaystyle 2\oint \: \vec{E}. \vec{ds} = \frac{q}{ \epsilon_{0}}$

$\displaystyle \implies2\oint \: E \times ds \times \cos( {0}^{ \circ} ) = \frac{q}{ \epsilon_{0}}$

$\displaystyle \implies2\oint \: E \times ds = \frac{q}{ \epsilon_{0}}$

$\displaystyle \implies 2E\oint ds = \frac{q}{ \epsilon_{0}}$

$\displaystyle \implies 2E \times s = \frac{q}{ \epsilon_{0}}$

$\displaystyle \implies E = \frac{q}{2s \epsilon_{0}}$

$\displaystyle \implies E = \frac{ \sigma}{2 \epsilon_{0}}$

Now, putting the values:

$\displaystyle \implies E = \frac{8.85 \times {10}^{ - 6} }{2 \times 8.85 \times {10}^{ - 12} }$

$\displaystyle \implies E = \frac{ {10}^{ 6} }{2}$

$\displaystyle \implies E = 5 \times {10}^{ 5} \: N/C$

So, field intensity (independent of distance) is 5 × 10⁵ N/C.