Question

# 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 intensity at a distance $$3r$$ from the centre is:

Solution

## $$V_{surface}=\cfrac{Kq}{r}$$$$V_{at\quad distance\quad 3r\quad from\quad centre}$$$$=\cfrac{Kq}{3r}$$Potential difference $$=V=\cfrac{Kq}{r}-\cfrac{kq}{3r}$$ $$=\cfrac{Kq}{r}.\cfrac{2}{3}=\cfrac{2Kq}{3r}$$Electric intensity (at distance $$3r$$ from the centre$$)=$$$$\Rightarrow E_{3r}=\cfrac{Kq}{qr^2}$$$$\because V=\cfrac{2}{3}\cfrac{Kq}{r}$$$$\cfrac{Kq}{r}=\cfrac{3V}{2}$$$$\therefore E_{3r}=\cfrac{3V}{2\times 9\times r}=\cfrac{V}{6r}$$$$E_{3r}=\cfrac{V}{6r}$$    $$[E_{3r}=$$ Electric intensity at distance $$3r$$ from the centre$$]$$PhysicsNCERTStandard XII

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