In the given diagram, there is a conducting sphere of radius r1 which is surrounded by a dielectric of relative permittivity ϵ r . If conducting sphere is given charge q , then the surface density of polarization charges, on the outer surface of the dielectric layer, is

Question

In the given diagram, there is a conducting sphere of radius r1 which is surrounded by a dielectric of relative permittivity ϵ r . If conducting sphere is given charge q , then the surface density of polarization charges, on the outer surface of the dielectric layer, is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-7jtm9qc27qqh.png" /> (A) (ε r q/4 π r22) (B) (q/4 π ε r r22) (C) ((ε r -1) q/4 π r22) (D) ((ε r -1) q/4 π ε r r22)

Tardigrade Admin · Accepted Answer

Induced charge =q(1-(1/ε r )) Surface density =(q(1-(1/ε F ))/4 π r22)=(q(ε r -1)/4 π ε r r22)

Q. In the given diagram, there is a conducting sphere of radius $r_{1}$ which is surrounded by a dielectric of relative permittivity $ϵ_{r}$ . If conducting sphere is given charge $q$ , then the surface density of polarization charges, on the outer surface of the dielectric layer, is

$\frac{ε _{r} q}{4 π r _{2}^{2}}$

$\frac{q}{4 π ε _{r} r _{2}^{2}}$

$\frac{( ε _{r} - 1 ) q}{4 π r _{2}^{2}}$

$\frac{( ε _{r} - 1 ) q}{4 π ε _{r} r _{2}^{2}}$

Induced charge $= q (1 - \frac{1}{ε _{r}})$
Surface density $= \frac{q ( 1 - \frac{1}{ε _{F}} )}{4 π r _{2}^{2}} = \frac{q ( ε _{r} - 1 )}{4 π ε _{r} r _{2}^{2}}$

Q. In the given diagram, there is a conducting sphere of radius r1​ which is surrounded by a dielectric of relative permittivity ϵr​ . If conducting sphere is given charge q , then the surface density of polarization charges, on the outer surface of the dielectric layer, is

4πr22​εr​q​

4πεr​r22​q​

4πr22​(εr​−1)q​

4πεr​r22​(εr​−1)q​