Question 772: Electrostats & Capacitance

Question

The electric field intensity near a conducting surface having uniform surface charge density ' $\sigma$ ' is given by ( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

Step 1: Concept
For an infinite sheet of charge (not necessarily conducting), the electric field is derived from
GaussÃ¢ÂÂs law
:
$ E = \frac{\sigma}{2 \varepsilon_0} $
on each side, and it is
normal
(perpendicular) to the surface.
Step 2: For a conductor
If the sheet is a
conductor
, all the charge resides on the surface.
Inside the conductor, the field must be
zero
.
Therefore, the entire field must appear
outside
the surface.
Thus, instead of being split into two equal parts ($ \frac{\sigma}{2\varepsilon_0} $ each side), the whole effect is
on one side only
.
So:
$ E = \frac{\sigma}{\varepsilon_0} $
directed
normal
to the surface (pointing outward if $\sigma > 0$, inward if $\sigma < 0$).
Step 3: Conclusion
Magnitude: $ 	frac{\sigma}{\varepsilon_0} $
Direction:
normal
to conducting surface
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Correct Answer: Option C
$ \frac{\sigma}{\varepsilon_0} \ 	ext{and is normal to the surface.} $

Answer

$\frac{\sigma}{\varepsilon_0}$ and is parallel to the surface

Answer

$\frac{2 \sigma}{\varepsilon_0}$ and is parallel to the surface

Answer

$\frac{\sigma}{\varepsilon_0}$ and is normal to the surface

Answer

$\frac{2 \sigma}{\varepsilon_0}$ and is normal to the surface

Electrostats & Capacitance

The electric field intensity near a conducting surface having uniform surface charge density ' \sigma ' is given by ( \varepsilon_0= permittivity of free space)

Choose the correct answer: