An electron is taken from point $A$ to point $B$ along the path $A B$ in a uniform electric field of intensity $E = 10 V m^{- 1}$ . Side $A B = 5 m$ , and side $B C = 3 m$ . Then, the amount of work done on the electron by us is :

50 eV

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40 eV

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-50 eV

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-40 eV

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Solution

The correct option is B 40 eV
$W_{A B} = W_{A C} + W_{C B}$
$W_{C B}$ should be zero, because in moving from $C$ to $B$ , we always move perpendicular to field. Hence, force applied by field and displacement will be at $90^{\circ}$ .
so work done in BC will be 0
$W_{A C} = - e (V_{C} - V_{A})$
$V_{C} - V_{A} = - E \times A C = - 10 \times 4 = - 40$
$∴ W_{A B} = 40 e J = 40 e V$

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An electron is taken from point A to point B along the path AB in a uniform electric field of intensity E=10Vm−1. Side AB=5m, and side BC=3m. Then, the amount of work done on the electron by us is :

The correct option is B 40 eVWAB=WAC+WCBWCB should be zero, because in moving from C to B, we always move perpendicular to field. Hence, force applied by field and displacement will be at 90∘.so work done in BC will be 0WAC=−e(VC−VA)VC−VA=−E×AC=−10×4=−40∴WAB=40eJ=40eV

An electron is taken from point $A$ to point $B$ along the path $A B$ in a uniform electric field of intensity $E = 10 V m^{- 1}$ . Side $A B = 5 m$ , and side $B C = 3 m$ . Then, the amount of work done on the electron by us is :