Question

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

• A. $50eV$
• B. $40eV$
• C. $-50eV$
• D. $-40eV$

Answer 1

The correct option is B $40eV$


As force due to electric field is a conservative force . The work done by this force is independent of the path followed by the electron.
Thus, $W_{AB}=W_{AC}+W_{CB}$

We know that $W.D=q(\Delta V$)
$⟹W.D=q(-\overrightarrow{E}.\overrightarrow{dr}$)
$⟹W.D=q(-\overrightarrow{E}\overrightarrow{dr}cos\theta$)

As the angle between displacement vector and electric field vector in CB is ${90}^0$.
$⟹W_{CB}=0$
$W_{AC}=-qErcos\left(0^0\right)$
$W_{AC}=-(-e)\times 10\times 4V$
$\therefore W_{AC}=W_{AB}=40eV$