The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $σ$ is
$σ / ϵ_{0}$
$σ / 2 ϵ_{0}$
$σ / ϵ_{0} r$
$σ / 2 ϵ_{0} r$

Question

The magnitude of the electric field on the surface of a sphere of radius r having a uniform surface charge density sigma issigma - epsilon-0-sigma - 2epsilon-0-sigma - epsilon-0-rsigma - 2epsilon-0-r

Toppr Admin · Accepted Answer

The magnitude of the electric field on the surface of radius -rCharge density -6Then-xA0-E-6-x3F5-0The electric field is independent of the surface radius-

The magnitude of the electric field on the surface of a sphere of radius r having a uniform surface charge density σ isσ/ϵ0σ/2ϵ0σ/ϵ0rσ/2ϵ0r

The magnitude of the electric field on the surface of radius =rCharge density =6Then, E=6ϵ0The electric field is independent of the surface radius.

The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $σ$ is
$σ / ϵ_{0}$
$σ / 2 ϵ_{0}$
$σ / ϵ_{0} r$
$σ / 2 ϵ_{0} r$

The magnitude of the electric field on the surface of radius $= r$
Charge density $= 6$
Then, $E = \frac{6}{ϵ_{0}}$
The electric field is independent of the surface radius.