$A, B, C, D, P$ 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:

Question

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Answer 1

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

$E_{n e t} = E_{x}^{2} + E_{y}^{2} = 152 V m^{- 1}$

Answer 2

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} = E_{x}^{2} + E_{y}^{2} = 152 V m^{- 1}