A small hole is cut into a charged hollow conductor of arbitrary shape as shown. If the local surface charge density near the hole is $$\sigma ,$$ then $$\vec { E }$$ inside the hole is
A
$$\dfrac { \sigma } { 2 \varepsilon _ { 0 } }$$ along outward normal
B
$$\dfrac { \sigma } { \varepsilon _ { 0 } }$$ along outward normal
C
$$\dfrac { \sigma } { \varepsilon _ { 0 } }$$ along inward normal
D
$$\dfrac { \sigma } { 2 \varepsilon _ { 0 } }$$ along inward normal
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Solution
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Correct option is A. $$\dfrac { \sigma } { \varepsilon _ { 0 } }$$ along outward normal
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A
hollow charged conductor has a tiny hole cut into its surface. Show
that the electric field in the hole is
,
where
is the unit vector in the outward
normal direction, and
is
the surface charge density near the hole.