An electron is taken from a point $A$ to point $B$ along the path $A B$ in a uniform electric field of intensity $E=10 \mathrm{Vm}^{-1}$, Side $A B=5 \mathrm{~m}$ and side $B C=3 \mathrm{~m} .$ Then, the amount of work done is (Mark answer in eV)

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Aicharya · Accepted Answer

$W_{A B}=W_{A C}+W_{C B}$ $W_{C B}$ should be zero, because in moving from $C$ to $B$, we always move perpendicular to the field. Hence, force applied by field and displacement will be at $90^{\circ}$. $W_{A C}=-e\left(V_{C}-V_{A}\right)$ $V_{C}-V_{A}=-E \times A C$ $=-10 \times 4=-40$ $W_{A C}=-e \times(-40)=40 e$ So $W_{A B}=40 e J=40 \mathrm{eV}$

An electron is taken from a point \(A\) to point \(B\) along the path \(A B\) in a uniform electric field of…

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