An electric field given by E→=(yi^+xj^)N/C .The work done in moving a 1C charge from r→A=(2i^+2j^)m to r→B=(4i^+j^)m is

W=∫dW=∫qE⋅dr¯=q∫(2,2)(4,1) (yi^+xj^)⋅(dx i^+dy j^)=q∫(ydx+xdy)=q∫ydx+∫xdy=q(yx+xy)=q2xy2,24,1=1×2×(4×1−2×2)=0

An electric field given by E→=(yi^+xj^)N/C .The work done in moving a 1C charge from r→A=(2i^+2j^)m to r→B=(4i^+j^)m is

W=∫dW=∫qE⋅dr¯=q∫(2,2)(4,1) (yi^+xj^)⋅(dx i^+dy j^)=q∫(ydx+xdy)=q∫ydx+∫xdy=q(yx+xy)=q2xy2,24,1=1×2×(4×1−2×2)=0

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An electric field given by $\vec{E} = (y \hat{i} + x \hat{j}) N / C$ .The work done in moving a 1C charge from ${\vec{r}}_{A} = (2 \hat{i} + 2 \hat{j}) m to {\vec{r}}_{B} = (4 \hat{i} + \hat{j}) m$ is

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zero

- 4J

+ 8J

+ 4J

answer is D.

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Detailed Solution

$\begin{array}{l} W = \int dW = \int q E \cdot d \bar{r} = q \int_{(2, 2)}^{(4, 1)} (y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \\ = q \int (ydx + xdy) = q [\int ydx + \int xdy] \\ = q (yx + xy) = {q 2 xy|}_{2, 2}^{4, 1} \\ = 1 \times 2 \times (4 \times 1 - 2 \times 2) = 0 \end{array}$