A solid conducting sphere of radius a has a net positive charge 2 Q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge -Q. The surface charge density on the inner and outer surfaces of the spherical shell will be

Question

A solid conducting sphere of radius a has a net positive charge 2 Q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge -Q. The surface charge density on the inner and outer surfaces of the spherical shell will be (A) -(2 Q/4 π b2), (Q/4 π c2) (B) -(Q/4 π b2), (Q/4 π c2) (C) 0, (Q/4 π c2) (D) None of the above

Tardigrade Admin · Accepted Answer

Surface charge density (σ)=( text Charge / text Surface area ) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0e40856c7be706509f0ac9aa1ed3cefe-.png" /> So σ text inner =(-2 Q/4 π b2) and σ text Outer =(Q/4 π c2)

$- \frac{2 Q}{4 π b ^{2}}, \frac{Q}{4 π c ^{2}}$

$- \frac{Q}{4 π b ^{2}}, \frac{Q}{4 π c ^{2}}$

$0, \frac{Q}{4 π c ^{2}}$

None of the above

Surface charge density $(σ) = \frac{Charge}{Surface area}$

So $σ_{inner} = \frac{- 2 Q}{4 π b ^{2}}$
and $σ_{Outer} = \frac{Q}{4 π c ^{2}}$

Q. A solid conducting sphere of radius a has a net positive charge 2Q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge −Q. The surface charge density on the inner and outer surfaces of the spherical shell will be

−4πb22Q​,4πc2Q​

−4πb2Q​,4πc2Q​

0,4πc2Q​

None of the above

Surface charge density (σ)= Surface area Charge ​ So σinner ​=4πb2−2Q​ and σOuter ​=4πc2Q​

$- \frac{2 Q}{4 π b ^{2}}, \frac{Q}{4 π c ^{2}}$

$- \frac{Q}{4 π b ^{2}}, \frac{Q}{4 π c ^{2}}$

$0, \frac{Q}{4 π c ^{2}}$

Surface charge density $(σ) = \frac{Charge}{Surface area}$

So $σ_{inner} = \frac{- 2 Q}{4 π b ^{2}}$
and $σ_{Outer} = \frac{Q}{4 π c ^{2}}$