A charge q is distributed over two concentric hollow spheres of radii r and r (r > r) such that the surface densities are equal. The potential at the common centre
Answers
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).
If you found this answer helpful, be sure to approve it as the right answer!
Explanation:
please mark me down as a BRAINLIEST for more answers please follow me.
Let the charge on inner sphere = q
Then, Charge on outer sphere = Q – q
Since, σ₁ = σ₂,
q / 4πr² = (Q-q) / 4πR²
q / r² = (Q - q) / R²
q = Qr² / (R² + r²)
and, Q - q = QR² / (R² + r²)
Hence, Potential at center O = V₀ = kq / r + k(Q-q)/R
= R [ q / r + (Q - q)/R] = [ Qr/(R² + r²) + QR/(R² + r²)
= QR ( R + r) / (R² + r²)
