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The graphical variation of the electric field due to a uniformly charged insulating solid sphere of radius R, with distance r from the centre O, is represented by

Asked in JEE Main 8th April 1st Shift 2023 · Shell and uniform solid sphere

Figure: Shell and uniform solid sphere
Answer: (4) E rises linearly from O to a maximum at r=R and then falls off as 1/(r²)

Step-by-step solution

Inside, a Gaussian sphere of radius r<R encloses qᵢₙ=Q(r³)/(R³), so E=(kQr)/(R³) — a straight line through the origin.

At r=R the two expressions meet at the maximum E=(kQ)/(R²).

Outside, the whole charge acts as if at the centre: E=(kQ)/(r²), falling off as 1/(r²).

So the graph rises linearly to a peak at r=R and then decays as an inverse square.

Why the other options are wrong

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