Physics · Electrostatics

JEE Main 2026 — 22 January, Morning Shift — Question 40

Electric field in a region is given by E⃗=Axi^+Byj^\vec{E}=A x \hat{i}+B y \hat{j}, where A=10 V/m2A=10 \mathrm{~V} / \mathrm{m}^{2} and B=5 V/m2B=5 \mathrm{~V} / \mathrm{m}^{2}. If the electric potential at a point (10,20)(10,20) is 500 V , then the electric potential at origin is ____\_\_\_\_ V.

  1. Option A:

    1000

  2. Option B:

    500

  3. Option C:

    2000

    Correct
  4. Option D:

    0

Answer: C

Step-by-step solution

E→=10xi^+5yj^\overrightarrow{\mathrm{E}}=10 x \hat{\mathrm{i}}+5 y \hat{\mathrm{j}} Vat (10,20)=500 V\mathrm{V}_{\text {at }(10,20)}=500 \mathrm{~V} ΔV−∫E→⋅dr→\Delta \mathrm{V}-\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}} 500−V0=−∫(0,0)(10,20)(10xi^+5yj^)⋅(dxi^+dyj^)500-V_{0}=-\int_{(0,0)}^{(10,20)}(10 x \hat{i}+5 y \hat{j}) \cdot(d x \hat{i}+d y \hat{j}) 500−V0=−[5x2+5y22](0,0)(10,20)500-V_{0}=-\left[5 x^{2}+\frac{5 y^{2}}{2}\right]_{(0,0)}^{(10,20)} V0−500=(500+5×4002)−(0−0)\mathrm{V}_{0}-500=\left(500+5 \times \frac{400}{2}\right)-(0-0) V0−500=500+1000\mathrm{V}_{0}-500=500+1000 V0=2000 V\mathrm{V}_{0}=2000 \mathrm{~V}

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
Electric field in a region is given by vec E =A x hat i +B y hat j … | JEE Main 2026 PYQ with Solution · DhiX AI