A charged particle (q = –8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB – VA?

-600 V

Explanation:

The expression for the relationship between work done and a charge is as follows;

Here, W is the work done, q is the charge and is the change in potential.

It is given in the problem that a charged particle is released from rest at point A.

The change in kinetic energy is equal to the work done. The kinetic energy is zero as the velocity of a charged particle is zero at point A.

The relation between change in kinetic energy and work done is as follows;

Here, is the final kinetic energy and is the initial kinetic energy.

According to question,

W= 4.8 J

Use the equation (1) to get the value of electric potential difference.

Put , W= 4.8 J and .

Therefore, the electric potential is -600 V.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)