Draw E-r and V-r graphs due to two charged spherical shells as shown in figure (along the line between `C` and `prop`)

Explanation

E-r graph
C to P `q_"in"=0implies E=0`
At M `E=(kq)/R^2` (radially outwards, say positive) `=E_0`(say)
At `N E=(kq)/((2R)^2)=(kq)/(4R^2)=E_0/e` (radialy outwards)
From M to N value will decrease from `E_0` to `E_0/4`
At `T E=(k(-2q+q))/((2R^2)^2)` (radially inwards)
`=-E_0/4`
From T to `oo` Value will decrease form `E_0` to `E_0/4`
At `T `E=(k(-2q+q))/((R)^2)` (radially inwards)
`=-E_0/4`
From `T` to `oo` Value changes from `-E_0/4 ` to zero.
V-r graph
From C to P Points are lying inside both the shells. Hence, potential due to both shell is constant
`:. V=(kq)/R-(k(2q))/(2R)=0`
From M to N potential of q will decrease. So, net value comes out to be negative. At N or T
`V=(kq)/(2R)-(k(2q))/(2R)`
`=-(kq)/(2R)=-V_0` (say)
From to `oo` value will chnge from `-V_0` to zero. The correct graph is a as shown below.