The correct option is
A 35KQ2RImagine that we assemble the sphere by building up a succession of thin spherical layers of infinitesimal thickness.
At each stage of the process, we gather a small amount of charge and put it in a thin shell from
r to
r+dr.
We continue the process until we arrive at the final radius
R.
If
Qr is the charge of the sphere when it has been built up to the radius
r, the work done in bringing a charge
dQ to it is
dW=dQ×Vr
⇒dW=QrdQ4πε0r...(1)
If the density of charge in the sphere is
ρ, the charge
Qr is
Qr=ρ×43πr3.
and the charge
dQ is
dQ=ρ×4πr2dr
Substituting the value in
(1), we get
dW=4πρ2r4dr3ϵ0.
The work done required to assemble the sphere is the integral of
dW from
r=0 to
r=R
∫dW=∫R04πρ2r4dr3ε0
⇒W=4πρ2R515ε0
Work done in terms of the total charge
Q=ρ43πR3 of the sphere,
W=4π(3Q4πR3)2R515ε0
⇒W=14πε035Q2R
Putting
14πε0=K, we get
W=35KQ2R
Hence, option (a) is the correct answer.
Why this question?
Note : The work done in gathering the charges together from infinity is called the self energy of the body.
It is also the work required to break the system apart [constant of binding energy] |