Let ρ(r) = Q r/π R^4 be the charge density distribution for a solid sphere of radius R and total charge Q . For a point ' p ' inside the sphere at dis - Physics MCQ AIEEE 2009 2026 | ExamTest.live

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Let ρ (r) = \frac{Q r}{π R ^{4}} be the charge density distribution for a solid sphere of radius R and total charge Q . For a point ' p' inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is:

Solution & Step-by-step Explanation

Using Gauss's Law for a point at distance r_{1} (r_{1} < R): \oint E \cdot d A = \frac{q _{e n c l ose d}}{ϵ _{0}} E (4 π r_{1}^{2}) = \frac{1}{ϵ _{0}} \int_{0}^{r_{1}} ρ (r) \cdot 4 π r^{2} d r E (4 π r_{1}^{2}) = \frac{1}{ϵ _{0}} \int_{0}^{r_{1}} \frac{Q r}{π R ^{4}} \cdot 4 π r^{2} d r E (4 π r_{1}^{2}) = \frac{4 π Q}{ϵ _{0} π R ^{4}} \int_{0}^{r_{1}} r^{3} d r = \frac{4 Q}{ϵ _{0} R ^{4}} [\frac{r ^{4}}{4}]_{0}^{r_{1}} = \frac{Q r _{1}^{4}}{ϵ _{0} R ^{4}} E = \frac{Q r _{1}^{4}}{4 π ϵ _{0} r _{1}^{2} R ^{4}} = \frac{Q r _{1}^{2}}{4 π ϵ _{0} R ^{4}}

Try it yourself before checking the explanation above.

Let ρ (r) = \frac{Q r}{π R ^{4}} be the charge density distribution for a solid sphere of radius R and total charge Q . For a point ' p' inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is:

0

\frac{Q}{4 π ϵ _{0} r _{1}^{2}}

\frac{Q r _{1}^{2}}{4 π ϵ _{0} R ^{4}}

\frac{Q r _{1}^{2}}{3 π ϵ _{0} R ^{4}}

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