Question

If a mass of 20g having charge 3.0mC moving with velocity 20$m{{s}^{-1}}$ enters a region of electric field of 80$N{{C}^{-1}}$ in the same direction as the velocity of mass, then the velocity of mass after 3s in the region will be
(A). 40$m{{s}^{-1}}$
(B). 44$m{{s}^{-1}}$
(C ). 56$m{{s}^{-1}}$
(D). 80$m{{s}^{-1}}$

Answer 1

Hint: Apply Newton’s second law of motion and then equate it with the force in an electric field by a moving charge. The force in electrostatics is the product of electric field and magnitude of charge. Then equating those two equations calculate the acceleration .

Complete step by step answer:
Given that,
m = 20g
q = 3.0mC
u = 20$m{{s}^{-1}}$
E = 80$N{{C}^{-1}}$
t = 3s
 As the mass is moving in the electric field then,
F = ma = qE
Then,
a= $\dfrac{qE}{m}$
Substituting the values of q, E and m we get,
$\Rightarrow a=\dfrac{3\times {{10}^{-3}}\times 80}{20\times {{10}^{-3}}}=12m{{s}^{-1}}$
By using,
$\Rightarrow v = u + at $
Substituting the values we get,
$\Rightarrow v = 20 + 12 \times 3 = 56 m{{s}^{-1}}$
The velocity of mass after 3s is $56m{{s}^{-1}}$.

Hence, option (C ) is correct.

Additional information:
Electric field is defined as the electric force per unit charge. The direction of the field is the same as the direction of the force. The electric field goes out from a positive charge and goes into a negative charge.

Note:
Here we should apply both electric force and Newton’s second law and comparing them we get a new equation of acceleration. And applying the equation of motion we will get the final velocity. Where ever we have initial and final velocity there is chance we can apply equations of motion