$A, B, C, D, P$ and $Q$ are points in uniform electric field. The potentials at these points are $V (A) = 2 v o l t$ . $V (P) = V (B) = V (D) = + 5 v o l t$ . $V (C) = 8 v o l t$ . The electric field at $P$ is:

10 V m^{- 1}

along

P Q

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15 \sqrt{2} V m^{- 1}

along

P A

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5 V m^{- 1}

along

P C

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5 V m^{- 1}

along

P A

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Solution

The correct option is C $15 \sqrt{2} V m^{- 1}$ along $P A$
Electric field $E = - \frac{d V}{d x} - \frac{d V}{d y} = - \frac{Δ V}{Δ x} - \frac{Δ V}{Δ y}$
$E_{x} = - \frac{V_{D} - V_{A}}{x_{D} - x_{A}} = - \frac{5 - 2}{0.2} = - 15 V / m$
$E_{y} = - \frac{V_{B} - V_{A}}{y_{B} - y_{A}} = - \frac{5 - 2}{0.2} = - 15 V / m$
$E_{x}$ and $E_{y}$ are same in magnitude and perpendicular to each other so resultant will be at angle of $π / 4$ or along with PA
$E_{n e t} = \sqrt{E_{x}^{2} + E_{y}^{2}} = 15 \sqrt{2} V m^{- 1}$

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A,B,C,D,P and Q are points in uniform electric field. The potentials at these points are V(A)=2 volt. V(P)=V(B)=V(D)=+5 volt. V(C)=8 volt. The electric field at P is:

$A, B, C, D, P$ and $Q$ are points in uniform electric field. The potentials at these points are $V (A) = 2 v o l t$ . $V (P) = V (B) = V (D) = + 5 v o l t$ . $V (C) = 8 v o l t$ . The electric field at $P$ is: