20) A hollow charged metal sphere has a radius \(r\). If the potential difference between its surface and a point at a distance \(3r\) from the centre is \(v\), then the electric field intensity at a distance \(3r\) is:
Solution:
For a hollow charged metal sphere, the potential at a distance \(x\) from the centre (outside the sphere) is given by: \[ V(x) = \frac{kQ}{x}, \] where \(k\) is the Coulomb constant and \(Q\) is the total charge.
The potential on the surface (at \(x = r\)) is: \[ V(r) = \frac{kQ}{r}. \]
The potential at a distance \(x = 3r\) is: \[ V(3r) = \frac{kQ}{3r}. \]
The given potential difference between the surface and the point at \(3r\) is: \[ v = V(r) - V(3r) = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r}. \]
Next, the electric field intensity at a distance \(x\) is related to the potential by: \[ E(x) = -\frac{dV}{dx}. \] However, for a point charge (or outside a charged sphere), the electric field is also given by: \[ E(x) = \frac{kQ}{x^2}. \]
At \(x = 3r\): \[ E(3r) = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2}. \]
We already found: \[ v = \frac{2kQ}{3r} \quad \Longrightarrow \quad kQ = \frac{3rv}{2}. \]
Substitute this into the expression for \(E(3r)\): \[ E(3r) = \frac{\frac{3rv}{2}}{9r^2} = \frac{3rv}{18r^2} = \frac{v}{6r}. \]
Therefore, the electric field intensity at \(3r\) is \(\frac{v}{6r}\), which corresponds to option C.
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