An uncharged thick spherical conducting shell is surrounding a chargeq at the center of the shell. Then charge +3q is placed on a point outside of the shell. When static equilibrium is reached, the total charges on the inner and outer surfaces of the shell are respectively:
Answers
W=q8πϵ0b(∫r=aσada+∫r=bσbda)
which come out to be zero as each integral is equal to q in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?
EDIT: I think I forgot the terms due to the potential of induced charges at the center.
W=q4πϵ0b(∫r=aσada+∫r=bσbda)+q24πϵ0b−q24πϵ0a
This comes out as
W=q24πϵ0(1b−1a)
Answer: -q , +q
Concept: Electric Flux, Guass Law
Given: Charge at the center of spherical shell = Q₁ = q
Charge at the outside of spherical shell = Q₂ = +3q
To Find: Charge at the inner surface of the conductor, Qi = ?
Charge at the outer surface of the conductor, Qo =?
Explanation:
Under static equilibrium, the total electric field, E = 0 .......(1)
As we know
Acc. to Gauss Theorem,
The net flux through a closed surface is directly proportional to the net charge in the volume enclosed by the closed surface.
ϕ = E*d
so, ϕ = 0 .......(2) (From eqn. (1))
Electric field, E is zero, then flux will also be zero.
Also ,
if ϕ is total flux
ϵ0 is electric constant,
The total electric charge Q enclosed by the surface is;
Q = ϕ*ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.
ϕ = 0 (From eqn. (2))
0= Q/ϵ0
0*ϵ0= Q
Q = 0
so, the total charge enclosed is zero.
Due to the Electric Induction, equal and opposite charge will be induced at the inner surface.
Therefore,
As charge at the center of spherical shell = Q₁ = q
Qi= -q
so, that the total charge enclosed will be zero.
Also,
Qo + Qi =0
so,
Qo =q
because in static equilibrium, the same charge remains at the outer surface of the spherical conductor., that is q.
Therefore, the charges on the inner and outer surfaces of the shell are ; -q, +q
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