Electrostatics

JEE Main level, single correct, medium

111 of 311

Question 111 of 311

Consider a sphere of radius RR which carries a uniform charge density ρ\rho . If a sphere of radius R2\dfrac{R}{2} is carved out of it, as shown, the ratio ∣EA⃗∣∣EB⃗∣\dfrac{\left|\vec{{E}_{A}}\right|}{\left|\vec{{E}_{B}}\right|} of magnitude of electric field EA⃗\vec{{E}_{A}} and EB⃗\vec{{E}_{B}} , respectively, at points AA and BB due to the remaining portion is:

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Answer: (B) 1834\dfrac{18}{34}

E=ρr3ϵ0E=\dfrac{\rho r}{3{\epsilon }_{0}}
EA=−ρR2(3ϵ0){E}_{A}=\dfrac{-\rho R}{2\left(3{\epsilon }_{0}\right)}
∣EA∣=ρR6ϵ0\left|{E}_{A}\right|=\dfrac{\rho R}{6{\epsilon }_{0}}
Electric filed at point B=EB=E1A+E2AB={E}_{B}={E}_{1A}+{E}_{2A}
E1A={E}_{1A}= Electric Filed Due to solid sphere of radius R at point B=ρR3ϵ0B=\dfrac{\rho R}{3{\epsilon }_{0}}
E2A={E}_{2A}= Electric Filed Due to solid sphere of radius R2\dfrac{R}{2} (which having charge density –ρ–\rho ) at point B
=−KQ′×49R2=−ρR54ϵ0=-\dfrac{K{Q}^{'}\times 4}{9{R}^{2}}=-\dfrac{\rho R}{54{\epsilon }_{0}}
EB=E1A+E2A=ρR3ϵ0−ρR54ϵ0=17ρR54ϵ0{E}_{B}={E}_{1A}+{E}_{2A}=\dfrac{\rho R}{3{\epsilon }_{0}}-\dfrac{\rho R}{54{\epsilon }_{0}}=\dfrac{17\rho R}{54{\epsilon }_{0}}
∣EA∣∣EB∣=917\dfrac{\left|{E}_{A}\right|}{\left|{E}_{B}\right|}=\dfrac{9}{17}

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