The correct option is
D By closing the switch the capacitance of the system increases
When the switch
S is open, no charge is present on inner shell.
Vinner shell=Vouter shell=kQ3R
So, Option
(a) is correct
Let the charge on the inner shell after closing the switch be
Q′
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1223142/original_85.png)
Since the inner shell is earthed, the net potential must be zero for it.
Vinner=0
The net potential at the inner shell is the sum of potential due to the inner shell itself and due to the outer shell.
⇒kQ′R+kQ3R=0⇒Q′=−Q3
So, Option
(b) and
(c) are correct.
Now from the definition of capacitance,
Ci=4πε0(3R)
The charge on inner shell is zero initially when switch is open, hence only capacitance of outer sphere is taken into account.
& Cf=4πε0(3R)+4πϵ0(1R−13R)
After the switch is closed the capacitance of both spheres add up, since they get connected in parallel.
⇒Cf>Ci
Hence, options (a) , (b) , (c) and (d) are the correct answers.