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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$$]$$


Physics
NCERT
Standard XII

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