A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as E = 2xi + 3y^2j + 4k N/C. The work done in moving the charge is _______ J.

The correct answer is Option 186. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q49 – JEE-MAIN 2026

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Home Physics Electric Charges and FieldsQuestion #20

A three coulomb charge moves from the point

(0, - 2, - 5)

to the point

(5, 1, 2)

in an electric field expressed as

\vec{E} = 2 x \hat{i} + 3 y^{2} \hat{j} + 4 \hat{k}

N/C. The work done in moving the charge is _______ J.

Enter Calculated Numerical Value:

Clue 1 of 3: Work-Energy Theorem

The work done by an electric field on a charge is given by the line integral of the electric force over the path.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify the formula for work done✦ Active

The work done $W$ by an electric field $\vec{E}$ on a charge $q$ moving from an initial point A to a final point B is given by the line integral:

W = q \int_{A}^{B} \vec{E} \cdot d \vec{l}

Given charge $q = 3$ C. The electric field is $\vec{E} = 2 x \hat{i} + 3 y^{2} \hat{j} + 4 \hat{k}$ N/C. The differential displacement vector is $d \vec{l} = d x \hat{i} + d y \hat{j} + d z \hat{k}$ . The initial point is $A = (0, - 2, - 5)$ and the final point is $B = (5, 1, 2)$ .

Step 2: Calculate the dot product and set up the integral○ Expand

First, calculate the dot product $\vec{E} \cdot d \vec{l}$ :

\vec{E} \cdot d \vec{l} = (2 x \hat{i} + 3 y^{2} \hat{j} + 4 \hat{k}) \cdot (d x \hat{i} + d y \hat{j} + d z \hat{k}) = 2 x d x + 3 y^{2} d y + 4 d z

Now, set up the integral for the work done:

W = q \int_{A}^{B} (2 x d x + 3 y^{2} d y + 4 d z)

W = q (\int_{x_{A}}^{x_{B}} 2 x d x + \int_{y_{A}}^{y_{B}} 3 y^{2} d y + \int_{z_{A}}^{z_{B}} 4 d z)

W = 3 (\int_{0}^{5} 2 x d x + \int_{- 2}^{1} 3 y^{2} d y + \int_{- 5}^{2} 4 d z)

Step 3: Evaluate the integrals and find the total work done○ Expand

Evaluate each definite integral:

\int_{0}^{5} 2 x d x = [x^{2}]_{0}^{5} = 5^{2} - 0^{2} = 25

\int_{- 2}^{1} 3 y^{2} d y = [y^{3}]_{- 2}^{1} = 1^{3} - (- 2)^{3} = 1 - (- 8) = 9

\int_{- 5}^{2} 4 d z = [4 z]_{- 5}^{2} = 4 (2) - 4 (- 5) = 8 - (- 20) = 28

Sum the results and multiply by the charge $q = 3$ C to find the total work done:

W = 3 (25 + 9 + 28) = 3 (62) = 186 J

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Physics Question 49 – JEE-MAIN 2026

Clue 1 of 3: Work-Energy Theorem

🧠 Full Solution Path