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A test charge q is moved without a acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure . Calculate the potential difference between A and C

$\begin{array}{1 1} 0.15 \\ 0.4 \\ 1.2\\ 4.7 \end{array} $

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A)
Solution :
$V_B=V_C$
$AB= 4\;cm (AB= \sqrt{45^2-9^2})$
$V_A-V_C= E \times (AB)$
$\quad= E \times 0.04 =0.04$
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