Electrostatics

JEE Main level, single correct, medium

154 of 311

Question 154 of 311

Two charges 7μC7 \mu \mathrm{C} and −2μC-2 \mu \mathrm{C} are placed at (−9,0,0)cm(-9,0,0) \mathrm{cm} and (9,0,0)cm(9,0,0) \mathrm{cm} respectively in an external field E=Ar2r^E=\frac{\mathrm{A}}{r^{2}} \hat{r}, where A=9×105 N/C.m2A=9 \times 10^{5} \mathrm{~N} / \mathrm{C}. \mathrm{m}^{2}. Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____\_\_\_\_ J.

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Answer: (A) 49.3

The total electrostatic energy UU of a system of two charges q1q_1 and q2q_2 in an external field is given by U=q1V(r1)+q2V(r2)+kq1q2r12U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}.
Given external electric field E=Ar2r^E = \frac{A}{r^2} \hat{r}, the potential V(r)V(r) is found by V(r)=−∫∞rE⋅dr=−∫∞rAr2dr=ArV(r) = -\int_{\infty}^{r} E \cdot dr = -\int_{\infty}^{r} \frac{A}{r^2} dr = \frac{A}{r}.
Parameters given:
q1=7μC=7×10−6Cq_1 = 7 \mu C = 7 \times 10^{-6} C at r1=9cm=0.09mr_1 = 9 cm = 0.09 m.
q2=−2μC=−2×10−6Cq_2 = -2 \mu C = -2 \times 10^{-6} C at r2=9cm=0.09mr_2 = 9 cm = 0.09 m.
Distance between charges r12=9−(−9)=18cm=0.18mr_{12} = 9 - (-9) = 18 cm = 0.18 m.
A=9×105N/C⋅m2A = 9 \times 10^5 N/C \cdot m^2.
Calculating individual terms:
q1V(r1)=q1Ar1=(7×10−6)9×1050.09=7×10−6×107=70Jq_1 V(r_1) = q_1 \frac{A}{r_1} = (7 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = 7 \times 10^{-6} \times 10^7 = 70 J.
q2V(r2)=q2Ar2=(−2×10−6)9×1050.09=−2×10−6×107=−20Jq_2 V(r_2) = q_2 \frac{A}{r_2} = (-2 \times 10^{-6}) \frac{9 \times 10^5}{0.09} = -2 \times 10^{-6} \times 10^7 = -20 J.
Interaction energy kq1q2r12=9×109×(7×10−6)×(−2×10−6)0.18=−126×10−30.18=−0.7J\frac{k q_1 q_2}{r_{12}} = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-2 \times 10^{-6})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 J.
Total energy U=70−20−0.7=49.3JU = 70 - 20 - 0.7 = 49.3 J.

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