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Physics · Electrostatics

JEE Main 2026 — 6 April, Morning Shift — Question 19

A three coulomb charge moves from the point (0,−2,−5)(0, -2, -5) to the point (5,1,2)(5, 1, 2) in an electric field expressed as E⃗=2xi^+3y2j^+4k^N/C\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k} \mathrm{N/C}. The work done in moving the charge is ______ J.

Answer: 186

Numerical answer — enter this value.

Step-by-step solution

Work = q∫E⃗⋅dl⃗=q[−ΔV]q \int \vec{E}\cdot d\vec{l} = q [ -\Delta V ] where V=−∫Exdx−∫Eydy−∫Ezdz=−x2−y3−4zV = -\int E_x dx - \int E_y dy - \int E_z dz = -x^2 - y^3 - 4z. Evaluate from (0,-2,-5) to (5,1,2): Vi=0−(−8)−(−20)=0+8+20=28V_i = 0 - (-8) - (-20) = 0+8+20=28? Wait careful: V=−∫2xdx=−x2V = -\int 2x dx = -x^2, −∫3y2dy=−y3-\int 3y^2 dy = -y^3, −∫4dz=−4z-\int 4 dz = -4z. So V = −x2−y3−4z-x^2 - y^3 - 4z. At initial (0,-2,-5): V_i = 0 - (-8) - (-20) = 0+8+20=28. At final (5,1,2): V_f = -25 - 1 - 8 = -34. ΔV = V_f - V_i = -34 - 28 = -62. Work = -q ΔV = -3×(-62)=186 J.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential

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