A uniform electric field of magnitude E = 100 | vedclass

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(D) The electric field vector $\vec{E}$ is directed along the line $y = x + 3$,which has a slope $m_1 = 1$. The unit vector in the direction of the el

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$0 \ V$

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(D) The electric field vector $\vec{E}$ is directed along the line $y = x + 3$,which has a slope $m_1 = 1$. The unit vector in the direction of the electric field is $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.Thus,$\vec{E} = 100 \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = 50\sqrt{2}(\hat{i} + \hat{j}) \ V/m$.The displacement vector $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (1 - 3)\hat{i} + (3 - 1)\hat{j} = -2\hat{i} + 2\hat{j}$.The potential difference is given by $V_A - V_B = \vec{E} \cdot \vec{r}_{AB}$.$V_A - V_B = [50\sqrt{2}(\hat{i} + \hat{j})] \cdot [-2\hat{i} + 2\hat{j}] = 50\sqrt{2}(-2 + 2) = 0 \ V$.Since the displacement vector is perpendicular to the electric field vector,the points $A$ and $B$ lie on the same equipotential line.

$A$ uniform electric field of magnitude $E = 100 \ V/m$ is directed along the line $y = x + 3$. Find the potential difference $V_A - V_B$ between points $A(3, 1)$ and $B(1, 3)$.

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