Question

A solid conducting sphere having a charge Q is surrounding by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the same two surfaces is(a) V(b) 2 V(c) 4 V(d) – 2 V

Answer 1

answer : option (A)

explanation : because the potential difference between solid sphere and hollow spherical shell depends on the radii of two spheres and charge on inner sphere. here charge and radius both remain constant. therefore, potential difference doesn't change.

so, potential difference between the surface is V.

mathematically,

initial potential difference, V = $\frac{KQ}{r_a}-\frac{KQ}{r_b}$ where $r_b > r_a$

Now when the shell is given a charge –3Q the potential at its surface and also inside will change by $V_0=\frac{-3KQ}{r_b}$

so now $V_{sphere}=K\left[\frac{Q}{r_a}-\frac{3Q}{r_b}\right]$

and $V_{shell}=K\left[\frac{Q}{r_b}-\frac{3Q}{r_b}\right]$

so, new potential difference, V' = $K\left[\frac{Q}{r_a}-\frac{Q}{r_b}\right]$

here it is clear that initial potential difference = final potential difference

Answer 2

As given solid sphere is surrounded by shell.

Solid sphere absolutely has mass and charge,

eventhough they have given as Shelluncharged due to induction shell also becomes charged.

Now moving to the sum

four formulas are involved in the sum, they are

potential of sphere on surface=1/4π£°{Q/r}

potential of sphere outside=1/4π£°{Q/r}

potential of shell on surface=1/4π£°{Q/r}

potential of shell inside=1/4π£°{Q/r}

So let us apply the data in the above formulas

first let us consider a point on surface of sphere

( please see the figure)

V1= 1/4π£°{Q/a} + 1/4π£°{Q/b}---------------------1

( potential of +. ( potential of shell

sphere surface). inside)

Here r=a and r=b

Now let us consider a point on the surface of shell

V2= 1/4π£°{Q/b}. +. 1/4π£°{Q/b}----------------2

( potential of +. (potential of shell

sphere outside). surface}

Here r=only b. because on seeing the figure when a point is considered on surface of shell

same distance b is for both shell and sphere.

Now potential difference

V1 - V2 = 1/4π£°{Q/a} - 1/4π£°{Q/b}--------------3

( given) V = Q/ {1/ a- 1/b}

4π£°

( here take Q/4π£° common and also you will get this answer by normal subtraction of equations 1 and 2) ( V1 - V2= V)

From equation 3 it is very clear that there is no role of the charge on shell.

So whatever, whether they add or remove -3Q there is no change in potential difference.

So potential remains same as V

Hope it is useful.

I ve mentioned all the points what my teacher said ,so explanation is big but the sum is actually 3 lines.

Have a marvellous day!!!!