JEE PYQ: Electrostatics Question 34
Question 34 - 2020 (09 Jan Shift 1)
Consider a sphere of radius $R$ which carries a uniform charge density $\rho$. If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown, the ratio $\frac{|\vec{E}_A|}{|\vec{E}_B|}$ of magnitude of electric field $\vec{E}_A$ and $\vec{E}_B$, respectively, at points A and B due to the remaining portion is:
(1) $\frac{21}{34}$
(2) $\frac{18}{34}$
(3) $\frac{17}{54}$
(4) $\frac{18}{54}$
Show Answer
Answer: (2)
Solution
Using superposition (full sphere minus carved sphere). At point A ($R/2$ from centre): $E_A = \frac{\rho(R/2)}{3\varepsilon_0} - 0 = \frac{\rho R}{6\varepsilon_0}$. At point B (on the surface, at distance $3R/2$ from carved sphere centre): $E_B = \frac{\rho R}{3\varepsilon_0} - \frac{\rho R}{54\varepsilon_0} = \frac{17\rho R}{54\varepsilon_0}$. $\frac{E_A}{E_B} = \frac{1/6}{17/54} = \frac{54}{102} = \frac{18}{34}$.
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