If two metallic spherical shell with radius R1 and R2 have same surface charge density then

Answer

Verified

Hint: The surface charge density for a spherical body is defined as the ratio of charge on the surface and the area of the surface. The potential is inversely proportional to the distance of separation of charge and the point where the potential is being measured.
Formula used: The surface charge density for a sphere is given as:$\sigma = \dfrac{Q}{\pi R^2}$.The potential due to a point charge at a distance d from it is given as:$V = \dfrac{Q}{4 \pi \epsilon_0 d}$

Complete answer:

We are given that the two spheres with different radii have the same surface charge density. The charge on the sphere 1 is therefore:$Q_1 = \sigma \pi {R_1}^2$and on sphere 2, the total charge is:$Q_2 = \sigma \pi {R_2}^2$ .Now, outside a charged sphere, we see that the sphere behaves as if the entire charge on it were located as a point charge at its centre. This is how we find the electric field of the sphere outside it. Therefore, both the spheres (at some distance from the sphere) will appear as if the entire charge were located at its centre. The potential in that case, solely depends on the charge that both spheres have. Since the charge depends on the square of radius, the ratio of potential at the same distance from the centre of two sphere will be:$\dfrac{V_1}{V_2} = \left( \dfrac{R_1}{R_2} \right)^2$

Therefore the correct answer is option (C).

Note:

Here, the point that we are measuring the distance from the centre of the two spheres is very important in determining the potential of the two spheres at a distance d from the centre of the two spheres. If the sphere is made of some charge, then the centre is where the charge is and distance from the centre has to be noted for the calculations. Also it has been assumed here that the spheres are conducting so all the charge resides on the surface.