A charge q is uniformly distributed over the surface of two concentric conducting spheres of radii r and r (r > r) such that surface charge densities are same for both spheres. Then potential at common centre of these spheres -
Answers
Answer:
Explanation:
By superposition princpiple, potential at the common centre is equal to algebraic sum of potentials at centre due to each sphere.
If we want the potential of a sphere, we need the radius (given) and the charge on it (which is what we should find now).
If the total charge is Q, then let’s assume charge of small sphere si q1, and large sphere is q2.
Thus Q = q1 + q2
It is given that the surface charge density is the same, thus:
(q1)/(4*pi*r^2) = (q2)/(4*pi*R^2).
Therefore,
q1 = (r^2)(q2)/(R^2)
But q1 + q2 = Q,
therefore,
q2 = Q(R^2)/(r^2 + R^2),
and similarly (from the same equation,
q1 = Q(r^2)/(r^2 + R^2).
Potential at common centre is now given as:
k(q1)/r + k(q2)/R.
Substituting previously found values, this becomes:
k(Q)(r+R)/(r^2 + R^2).
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