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(C) The potential difference between two points in a uniform electric field is given by the formula $\Delta V = -\vec{E} \cdot \vec{d}$,where $\vec{d}

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$-10 \ V$

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(C) The potential difference between two points in a uniform electric field is given by the formula $\Delta V = -\vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector from the initial point to the final point.For the potential difference $V_A - V_B$,the displacement vector is $\vec{BA}$.The magnitude of the displacement is $d = |\vec{BA}| = 2 \ m$.The angle $	heta$ between the electric field $\vec{E}$ and the displacement vector $\vec{BA}$ is $60^\circ$ (as it is an equilateral triangle).Therefore,$V_A - V_B = -E d \cos(	heta)$.Substituting the given values: $V_A - V_B = -(10 \ N \ C^{-1}) 	imes (2 \ m) 	imes \cos(60^\circ)$.$V_A - V_B = -20 	imes 0.5 = -10 \ V$.

An imaginary equilateral triangle $ABC$ of side length $2 \ m$ is placed in a uniform electric field $\vec{E} = 10 \ N \ C^{-1}$ as shown. Then,$V_A - V_B =$

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