A uniform electric field vector E exists alon | vedclass

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(A) The electric potential at point $A$ is $V_A$.Since the electric field $E$ is uniform and directed towards the right,the potential decreases in the

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$q(V_A - Ex)$

Vedclass · Answer

(A) The electric potential at point $A$ is $V_A$.Since the electric field $E$ is uniform and directed towards the right,the potential decreases in the direction of the electric field.The horizontal distance between $A$ and $B$ is $x$. Therefore,the potential difference between $A$ and $B$ is $\Delta V = V_B - V_A = -E \cdot x$.Thus,the potential at $B$ is $V_B = V_A - Ex$.The potential energy $U$ of a charge $q$ at a point with potential $V$ is given by $U = qV$.Therefore,the potential energy of the charge at point $B$ is $U_B = q V_B = q(V_A - Ex)$.

$A$ uniform electric field vector $E$ exists along the horizontal direction as shown. The electric potential at $A$ is $V_A$. $A$ small point charge $q$ is slowly moved from $A$ to $B$ along the curved path as shown. The potential energy of the charge when it is at point $B$ is

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