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(A) Given,electric field $E = 500 \ Vm^{-1}$ directed at $	heta = 30^{\circ}$ with the positive $X$-axis.The electric field vector is $\vec{E} = E(\c

Vedclass · Accepted Answer

$-250(3 \sqrt{3} + 5) \ V$

Vedclass · Answer

(A) Given,electric field $E = 500 \ Vm^{-1}$ directed at $	heta = 30^{\circ}$ with the positive $X$-axis.The electric field vector is $\vec{E} = E(\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = 500 \left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}ight) = (250\sqrt{3} \hat{i} + 250 \hat{j}) \ Vm^{-1}$.The coordinates of point $A$ are $(-3, 0) \ m$ and point $B$ are $(0, 5) \ m$.The displacement vector from $A$ to $B$ is $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (0 - (-3)) \hat{i} + (5 - 0) \hat{j} = (3 \hat{i} + 5 \hat{j}) \ m$.The potential difference is given by $\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} = -\vec{E} \cdot \vec{r}_{AB}$.Substituting the values:$V_B - V_A = -(250\sqrt{3} \hat{i} + 250 \hat{j}) \cdot (3 \hat{i} + 5 \hat{j})$$V_B - V_A = -(250\sqrt{3} 	imes 3 + 250 	imes 5)$$V_B - V_A = -250(3\sqrt{3} + 5) \ V$.

$A$ uniform electric field of $500 \ Vm^{-1}$ is directed at $30^{\circ}$ with the positive $X$-axis as shown in the figure. The potential difference $(V_B - V_A)$ if $OA = 3 \ m$ and $OB = 5 \ m$ is

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