Three concentric conducting spherical shells of radii `R, 2R` and `3R` carry charges `Q, - 2Q` and `3Q`, respectively.

a. Find the electric potential at `r=R` and `r=3R` where r is the radial distance from the centre. ltbr. b. Compute the electric field at `r=5/2R`
`c. compute the total electrostatic energy stored in the system.
The inner shell is now connected to the external one by a conducting wire, passing through a very small hole in the middle shell. ltbr. Compute the charges on the spheres of radii R and `3R`.
e. Compute the electric field at `r=5/2R`.

Answer

A, B, C

Explanation

a. At `r=R`,
`V=1/(4piepsilon_0)[Q/R-(2Q)/(2R)+(3Q)/(3R)]`
`=(1/(4piepsilon_0))(Q/R)`
`At r=3R`
`V=1/(4piepsilon_0)[Q/(3R)-(2Q)/(3R)+(3Q)/(3R)]=Q/(6piepsilon_0R)`
b. `E=(1/(4piepsilon_0)(q_("in")/r^2)`
`=(1/(4piepsilon_0))[(Q-2Q)/((5R//2))]=(-Q)/(25piepsilon_0R^2)`
Minus sign implies that this electric field is radialy towards centre.
c.
`U_T=U_1+U_2+U_3`
`where U_1=1/2 Q^2/C_1`
`where C_1=(4piepsilon_0)//(1/R-1/(2R))`
`U_2=1/2 Q^2/C_2`
where, `C_2=4piepsilon_0//(1/(2R)-1/(3R))`
and `U_3=1/2((2Q)^2)/C_3`
where `C_3=4piepsilon_0//(1/(3R)-1/oo)` d.
total charge on `(A+C)` is `3Q+Q=3Q`
Now, `V_A=V_C`
`:. (kq)/R-(k(2Q))/(2R)=(k(4Q-q))/(3R)`
`=(k(Q-2Q+3Q))/(3R)`
solving this equation we get
`q=Q/2`
Now, `q_A=q=Q/2`
`q_C=4Q-q=7/2Q`
e. `E=1/(4piepsilon_0).q_("in")/r^2=1/(4piepsilon_0)[(q-2Q)/((5R/2)^2)]=(-3Q)/(50piepslon_0R^2)` Minus sign indicates that electric field is radially inwards.