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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2020
  • Hard
  • MCQ

Consider a sphere of radius \(\mathrm{R}\) which carries a uniform charge density \(\rho .\) If a sphere of radius \(\frac{\mathrm{R}}{2}\) is carved out of it, as shown, the ratio \(\frac{\left|\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right|}{\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|}\) of magnitude of electric field \(\overrightarrow{\mathrm{E}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{E}}_{\mathrm{B}}\) respectively, at points \(\mathrm{A}\) and \(\mathrm{B}\) due to the remaining portion is

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Answer & Solution

Correct Answer

(D) \(\frac{18}{34}\)

Step-by-step Solution

Detailed explanation

Fill the empty space with \(+\rho\) and \(-\rho\) charge density.

\(\left|\mathrm{E}_{\mathrm{A}}\right|=0+\frac{\operatorname{k\rho} \cdot \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3}}{\left(\frac{\mathrm{R}}{2}\right)^{2}}=\operatorname{k\rho} \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)\)

\(\left|\mathrm{E}_{\mathrm{B}}\right|=\frac{\mathrm{k} \rho \cdot \frac{4}{3} \pi \mathrm{R}^{3}}{\mathrm{R}^{2}}-\frac{\mathrm{k} \rho \cdot \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3}}{\left(\frac{3 \mathrm{R}}{2}\right)^{2}}\)

\(=\operatorname{k\rho} \frac{4}{3} \pi \mathrm{R}-\mathrm{k} \rho \frac{4}{3} \pi \frac{\mathrm{R}}{18}=\mathrm{k} \rho \cdot \frac{4}{3} \pi\left(\frac{17 \mathrm{R}}{18}\right)\)

\(\frac{E_{A}}{E_{B}}=\frac{9}{17}=\frac{18}{34}\)

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