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Question

(B) The relationship between electric field $\vec E$ and electric potential $V$ is given by $\vec E = -
abla V$,which implies $dV = -\vec E \cdot d\v

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The relationship between electric field $\vec E$ and electric potential $V$ is given by $\vec E = -
abla V$,which implies $dV = -\vec E \cdot d\vec r$.Integrating from the origin $(0, 0, 0)$ to the point $(1, 1, 1)$:$V(1, 1, 1) - V(0, 0, 0) = -\int_{(0,0,0)}^{(1,1,1)} (2x\hat i + y^2\hat j) \cdot (dx\hat i + dy\hat j + dz\hat k)$.$V(1, 1, 1) - 2 = -\left[ \int_{0}^{1} 2x \, dx + \int_{0}^{1} y^2 \, dy ight]$.$V(1, 1, 1) - 2 = -\left[ x^2 ight]_{0}^{1} - \left[ \frac{y^3}{3} ight]_{0}^{1}$.$V(1, 1, 1) - 2 = -(1 - 0) - (1/3 - 0) = -1 - 1/3 = -4/3$.$V(1, 1, 1) = 2 - 4/3 = 2/3 \, V$.

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

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