Electrostatics

JEE Main level, numerical answer, medium

149 of 311

Question 149 of 311

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^\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k} N/C. The work done in moving the charge is _______ J.

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Answer: 186

The force experienced by the charge in the electric field is given by F⃗=qE⃗\vec{F} = q\vec{E}.

Given q=3q = 3 C and E⃗=2xi^+3y2j^+4k^\vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k} N/C, the force is:
F⃗=3(2xi^+3y2j^+4k^)=6xi^+9y2j^+12k^\vec{F} = 3(2x\hat{i} + 3y^2\hat{j} + 4\hat{k}) = 6x\hat{i} + 9y^2\hat{j} + 12\hat{k} N

The work done by the electric field in moving the charge is:
W=∫F⃗⋅dr⃗=∫x1x2Fxdx+∫y1y2Fydy+∫z1z2FzdzW = \int \vec{F} \cdot d\vec{r} = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz

Substituting the limits from the initial point (0,−2,−5)(0, -2, -5) to the final point (5,1,2)(5, 1, 2):
W=∫056xdx+∫−219y2dy+∫−5212dzW = \int_{0}^{5} 6x dx + \int_{-2}^{1} 9y^2 dy + \int_{-5}^{2} 12 dz

Evaluating each integral separately:
∫056xdx=[3x2]05=3(52−02)=75\int_{0}^{5} 6x dx = \left[ 3x^2 \right]_{0}^{5} = 3(5^2 - 0^2) = 75 J

∫−219y2dy=[3y3]−21=3(13−(−2)3)=3(1−(−8))=3(9)=27\int_{-2}^{1} 9y^2 dy = \left[ 3y^3 \right]_{-2}^{1} = 3(1^3 - (-2)^3) = 3(1 - (-8)) = 3(9) = 27 J

∫−5212dz=[12z]−52=12(2−(−5))=12(7)=84\int_{-5}^{2} 12 dz = \left[ 12z \right]_{-5}^{2} = 12(2 - (-5)) = 12(7) = 84 J

Total work done W=75+27+84=186W = 75 + 27 + 84 = 186 J.

Answer: 186186

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