In a region,if the electric field is defined | vedclass

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(B) The potential difference $V_{AB} = V_B - V_A$ is given by the line integral of the electric field: $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The potential difference $V_{AB} = V_B - V_A$ is given by the line integral of the electric field: $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.Given $\vec{E} = \hat{i} + 2\hat{j} + \hat{k}$ and the displacement vector $d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}$.The points are $A(0, 0, 0)$ and $B(2, 3, 4)$.$V_A - V_B = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{(0,0,0)}^{(2,3,4)} (\hat{i} + 2\hat{j} + \hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k})$.$V_A - V_B = \int_{0}^{2} dx + \int_{0}^{3} 2dy + \int_{0}^{4} dz$.$V_A - V_B = [x]_0^2 + 2[y]_0^3 + [z]_0^4 = 2 + 2(3) + 4 = 2 + 6 + 4 = 12 	ext{ V}$.Since the question asks for the potential difference between $A$ and $B$,we consider the magnitude or the path integral $V_A - V_B = 12 	ext{ V}$.

In a region,if the electric field is defined as $\vec{E} = (\hat{i} + 2\hat{j} + \hat{k}) \text{ V/m}$,then the potential difference between two points $A(0, 0, 0)$ and $B(2, 3, 4)$ in that region is ...... $V$.

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