Question

A small particle has charge -2.00 uC and mass 1.50×10-4 kg. It moves from point A, where the electric potential is V(A) = 200 V, to point B, where the electric potential V(B) = 920 V is greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 4.30 m/s at point A.
1.)What is its speed at point B?
2.)And is it moving faster or slower at B than at A?

Answer 1

Answer:

$v_b=6.13 m/s.$.

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

$T.E=qV_a+\dfrac{1}{2}mv_a^2=qV_b+\dfrac{1}{2}mv_b^2\\\dfrac{1}{2}mv_b^2=\dfrac{1}{2}mv_a^2+q(V_a-V_b)$

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get,  $v_b=6.13 m/s.$.

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.