Question

A uniform electric field of strength →E exists in a region . An electron enters a point A with velocity v as shown . It moves through the electric field and reaches at point B . Velocity of particle at B is 2v at 30∘ with x - axis . Then electric field →E is

Answer 1

(The diagram of the question is attached in the answer tab )

Given :

Velocity of the electron at point A= v

Velocity of the electron at point B = 2v

To Find :

Electric field = ?

Solution :

• By using Work-Energy theorem, Work done = Change in K.E

                  $W=\Delta K.E$

                  $F\times d = \Delta K.E$

• As the particle considered here is electron it will move away from the electric field. Also the velocity in x-axis is changing the displacement in x-axis direction is considered

                  $(eE)a=\frac{1}{2} m(4 v^{2} -v^{2} )$

                  $E=-\frac{3mv^{2}}{2ea}$

The magnitude of the electric field is   $E=-\frac{3mv^{2}}{2ea}$