A charge q is placed in front of a neutral metallic sphere. The magnitude of electric field at the centre of sphere due to the induced charge on the surface of sphere is

Question

A charge q is placed in front of a neutral metallic sphere. The magnitude of electric field at the centre of sphere due to the induced charge on the surface of sphere is (A) E = (1/4 π ϵo) (q/R2) (B) E = (1/4 π ϵo) (q/R) (C) E = (1/4 π ϵo) (q/d2) (D) Zero

Tardigrade Admin · Accepted Answer

Net electric field inside metal is zero. Esphere+ vecEq=0 Esphere= - vecEq=(1/4πεo)(q/d2) |Esphere|=(1/4πεo)(q/d2)

Q. A charge q is placed in front of a neutral metallic sphere. The magnitude of electric field at the centre of sphere due to the induced charge on the surface of sphere is

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{R ^{2}}$

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{R}$

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{d ^{2}}$

$Z ero$

Net electric field inside metal is zero.
$E_{s p h ere} + E_{q} = 0$
$E_{s p h ere} = - E_{q} = \frac{1}{4 π ε _{o}} \frac{q}{d ^{2}}$
$∣ ∣ E_{s p h ere} ∣ ∣ = \frac{1}{4 π ε _{o}} \frac{q}{d ^{2}}$

Q. A charge q is placed in front of a neutral metallic sphere. The magnitude of electric field at the centre of sphere due to the induced charge on the surface of sphere is

E=4πϵo​1​R2q​

E=4πϵo​1​Rq​

E=4πϵo​1​d2q​

Zero

Net electric field inside metal is zero. Esphere​+Eq​​=0 Esphere​=−Eq​​=4πεo​1​d2q​ ∣∣​Esphere​∣∣​=4πεo​1​d2q​

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{R ^{2}}$

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{R}$

$E = \frac{1}{4 π ϵ _{o}} \frac{q}{d ^{2}}$

$Z ero$

Net electric field inside metal is zero.
$E_{s p h ere} + E_{q} = 0$
$E_{s p h ere} = - E_{q} = \frac{1}{4 π ε _{o}} \frac{q}{d ^{2}}$
$∣ ∣ E_{s p h ere} ∣ ∣ = \frac{1}{4 π ε _{o}} \frac{q}{d ^{2}}$