A conductor with a cavity is charged positively and its surface charge density is σ . If E and V represent the electric field and potential, then inside the cavity

Question

A conductor with a cavity is charged positively and its surface charge density is σ . If E and V represent the electric field and potential, then inside the cavity (A) E=0 and V=0 (B) E=0 and σ = constant (C) V=0 and σ = constant (D) E=0 and V= textconstant

Tardigrade Admin · Accepted Answer

Surface charge density =σ Inside the surface of the conductor, the electric field is zero. ⇒ Ei n s i d e=0 Therefore, Vi n s i d e=Vs u r f a c e The potential inside the surface of charged sphere is constant textHence,E=0, V= constant.

Q. A conductor with a cavity is charged positively and its surface charge density is $σ$ . If $E$ and $V$ represent the electric field and potential, then inside the cavity

$E = 0$ and $V = 0$

$E = 0$ and $σ =$ constant

$V = 0$ and $σ =$ constant

$E = 0$ and $V = constant$

Surface charge density $= σ$
Inside the surface of the conductor, the electric field is zero.
$\Rightarrow E_{in s i d e} = 0$
Therefore, $V_{in s i d e} = V_{s u r f a ce}$
The potential inside the surface of charged sphere is constant
$Hence, E = 0, V =$ constant.

Q. A conductor with a cavity is charged positively and its surface charge density is σ . If E and V represent the electric field and potential, then inside the cavity

E=0 and V=0

E=0 and σ= constant

V=0 and σ= constant

E=0 and V=constant