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(B) The electric field $\vec{E}$ is uniform and directed along the $+X$-direction.Points $P(0,0)$ and $R(0,2)$ lie on the same plane perpendicular to

Vedclass · Accepted Answer

$V_P = V_R, V_P > V_Q$

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(B) The electric field $\vec{E}$ is uniform and directed along the $+X$-direction.Points $P(0,0)$ and $R(0,2)$ lie on the same plane perpendicular to the electric field lines (the $YZ$-plane). Therefore,$P$ and $R$ lie on the same equipotential surface,which implies $V_P = V_R$.Electric potential decreases in the direction of the electric field. Since point $Q(2,0)$ is further along the $+X$-direction than point $P(0,0)$,the potential at $P$ must be greater than the potential at $Q$,i.e.,$V_P > V_Q$.Combining these,we get $V_P = V_R$ and $V_P > V_Q$.

$A$ uniform electric field is prevailing in the $X$-direction in a certain region. The coordinates of points $P$,$Q$,and $R$ are $(0,0)$,$(2,0)$,and $(0,2)$ respectively. Which of the following alternatives is true for the potentials at these points?

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