In a uniformly charged sphere of total charge | vedclass

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Question

(B) For a uniformly charged sphere of total charge $Q$ and radius $R$:$1$. Inside the sphere $(r < R)$: The electric field is given by $E = \frac{1}{4

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(B) For a uniformly charged sphere of total charge $Q$ and radius $R$:$1$. Inside the sphere $(r $2$. At the surface $(r = R)$: The electric field is maximum,$E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}$.$3$. Outside the sphere $(r > R)$: The sphere behaves as a point charge at the center,so $E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}$. This shows that $E \propto \frac{1}{r^2}$,meaning the electric field decreases inversely with the square of the distance.Comparing this behavior with the given options,the graph that shows a linear increase for $r  R$ is the correct one.

In a uniformly charged sphere of total charge $Q$ and radius $R$,the electric field $E$ is plotted as a function of distance $r$ from the center. The graph which would correspond to the above is:

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