Physics · Electrostatics

JEE Main 2026 — 23 January, Evening Shift — Question 31

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}}{\mathrm{r}^{2}} \hat{\mathrm{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.

  1. Option A:

    1.4

  2. Option B:

    −90.7-90.7

  3. Option C:

    49.3

    Correct
  4. Option D:

    24.3

Answer: C

Step-by-step solution

↦q1=7μC(−9,0,0)r,(0,0,0)r(9,0,0)q2=−2μC\underset{(-9,0,0)}{\stackrel{\mathrm{q}_{1}=7 \mu \mathrm{C}}{\mapsto}} \mathrm{r} \underset{(0,0,0)}{,} \underset{(9,0,0)}{r} \mathrm{q}_{2}=-2 \mu \mathrm{C} dV=−E→⋅dr→\mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dr}} ∫0vdV=−∫∞rAr2dr\int_{0}^{\mathrm{v}} \mathrm{dV}=-\int_{\infty}^{\mathrm{r}} \frac{\mathrm{A}}{\mathrm{r}^{2}} \mathrm{dr} V=−[−Ar2]∞r⇒V=Ar\mathrm{V}=-\left[\frac{-\mathrm{A}}{\mathrm{r}^{2}}\right]_{\infty}^{\mathrm{r}} \Rightarrow \mathrm{V}=\frac{\mathrm{A}}{\mathrm{r}} U=Uself +Uinteraction \mathrm{U}=\mathrm{U}_{\text {self }}+\mathrm{U}_{\text {interaction }} =q1v1=q2v2+kq1q22r=\mathrm{q}_{1} \mathrm{v}_{1}=\mathrm{q}_{2} \mathrm{v}_{2}+\frac{\mathrm{kq}_{1} \mathrm{q}_{2}}{2 \mathrm{r}} =7×10−6 A9×10−2−2×10−6 A9×10−2=7 \times 10^{-6} \frac{\mathrm{~A}}{9 \times 10^{-2}}-2 \times 10^{-6} \frac{\mathrm{~A}}{9 \times 10^{-2}} −9×109×14×10−122×9×10−2-\frac{9 \times 10^{9} \times 14 \times 10^{-12}}{2 \times 9 \times 10^{-2}} =5×10−6×9×1059×10−2−7×10−1=\frac{5 \times 10^{-6} \times 9 \times 10^{5}}{9 \times 10^{-2}}-7 \times 10^{-1} =50−0.7=50-0.7 =49.3 J=49.3 \mathrm{~J}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential
Two charges 7 μ C and -2 μ C are placed at (-9,0,0) cm and (9,0,0) cm… | JEE Main 2026 PYQ with Solution · DhiX AI