A spherical charged conductor has a as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

Question

A spherical charged conductor has a as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ? (A)  (E/4) (B)  (E/2) (C) E (D) 2 E

Tardigrade Admin · Accepted Answer

E = (1/4 π ε0) (Q/R2) = (1/4 π ε0) (4 π R2 σ/R2) = (σ/ε0) It is independent of radius and depends on σ

Q. A spherical charged conductor has a as the surface density of charge. The electric field on its surface is $E$ . If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

$\frac{E}{4}$

$\frac{E}{2}$

E

2 E

$E = \frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}} = \frac{1}{4 π ε _{0}} \frac{4 π R ^{2} σ}{R ^{2}} = \frac{σ}{ε _{0}}$
It is independent of radius and depends on $σ$

Q. A spherical charged conductor has a as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

4E​

2E​

E

2 E

E=4πε0​1​R2Q​=4πε0​1​R24πR2σ​=ε0​σ​ It is independent of radius and depends on σ

Q. A spherical charged conductor has a as the surface density of charge. The electric field on its surface is $E$ . If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

$\frac{E}{4}$

$\frac{E}{2}$

$E = \frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}} = \frac{1}{4 π ε _{0}} \frac{4 π R ^{2} σ}{R ^{2}} = \frac{σ}{ε _{0}}$
It is independent of radius and depends on $σ$