A uniform electric field of magnitude $100 V/m$ in space is directed along the line $y = 3 + x$ . Find the potential difference (in $V$ ) between points $A (3, 1)$ and $B (1, 3)$ .

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Solution

Given :-
Electric field is along line $y = 3 + x$ , so, $m_{1} = tan θ = 1$ .

Also, slope of line joining points $A (3, 1)$ and $B (1, 3)$ ,
$m_{2} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 - 1}{1 - 3} = - 1$

As, $m_{1} . m_{2} = - 1$ , the line $A B$ and electric field are perpendicular. Therefore, $Δ V = \to E . Δ \to r = 0$

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A uniform electric field of magnitude 100 V/m in space is directed along the line y=3+x. Find the potential difference (in V) between points A(3,1) and B(1,3).

Given :- Electric field is along line y=3+x, so, m1=tanθ=1. Also, slope of line joining points A(3,1) and B(1,3), m2=y2−y1x2−x1=3−11−3=−1 As, m1.m2=−1, the line AB and electric field are perpendicular. Therefore, ΔV=→E.Δ→r=0

A uniform electric field of magnitude $100 V/m$ in space is directed along the line $y = 3 + x$ . Find the potential difference (in $V$ ) between points $A (3, 1)$ and $B (1, 3)$ .