Question

A particle having mass 1g and electric charge \[{10^{ - 8}}C\] travels from a point A having electric potential 600V to point B having zero potential. What would be the change in its kinetic energy?
A) $ - 6 \times {10^{ - 6}}erg$
B) $ - 6 \times {10^{ - 6}}J$
C) $6 \times {10^{ - 6}}J$
D) $6 \times {10^{ - 6}}erg$

Answer 1

Hint: As electric force is a conservative force, any particle moving under it will conserve its total mechanical energy. So the change in kinetic energy will be the same as the change in mechanical energy. So, try to find the amount of change in potential energy to get the answer.

Formula used:
Conservation of mechanical energy:
$\Delta K = - \Delta P$ (1)
Where,
$\Delta K$ is the change in Kinetic energy of the particle,
$\Delta P$ is the change in the Potential energy of the particle.
Change in potential energy:
$\Delta P = Q \times \Delta V$ (2)
Where,
Q is the electric charge of the particle,
$\Delta V$ is the difference in potential between the two points.

Complete step by step solution:
Given:
Electric charge of the particle \[Q = {10^{ - 8}}C\].
Potential at $A$ is ${V_A} = 600V$and at B is ${V_B} = 0V$.
Hence, $\Delta V = {V_B} - {V_A} = (0 - 600)V = - 600V$.

To find: Change in kinetic energy i.e. $\Delta K$.

Step 1
Substitute the given values of Q and $\Delta V$in eq(2) to get $\Delta P$:
$\Delta P = {10^{ - 8}}C \times ( - 600)V = - 6 \times {10^{ - 6}}J$
Here, the negative sign indicates that the potential energy of the particle will decrease.

Step 2
Now, using the obtained value of $\Delta P$ in eq(1) we get the value of $\Delta K$ as:
$\Delta K = - \Delta P = - ( - 6 \times {10^{ - 6}}J) = 6 \times {10^{ - 6}}J$
So, the Kinetic energy of the particle will increase.

$\therefore$ Change in kinetic energy will be $6 \times {10^{ - 6}}J$. Hence option (C) is the correct answer.

Note:
Many students make a common mistake while finding $\Delta V$. Potential difference is always calculated as (final – initial) not (higher potential – lower potential). In this case, the particle is moving from point A to point B. Hence $\Delta V = {V_B} - {V_A}$ not the reverse. Always be careful while finding the potential difference.