Question

A stationary body of mass 5 g carries a charge of 5 $\mu$C. The potential difference with which it should be accelerated to acquire a speed of 10 m/s is:

• A. 4 kV
• B. 25 kV
• C. 50 kV
• D. 40 kV
• E. 2 kV

Hint:

The kinetic energy gained by a charged particle in an electric field is equal to the work done, which can be calculated using \( {K.E.} = qV \).

Answer 1

The Correct Option is C

Step 1: The kinetic energy gained by the body when it is accelerated through a potential difference \( V \) is given by: \[ K.E = \frac{1}{2} m v^2 = qV \] 
where: 
- \( m = 5 \, {g} = 5 \times 10^{-3} \, {kg} \), 
- \( v = 10 \, {m/s} \), 
- \( q = 5 \, \mu C = 5 \times 10^{-6} \, {C} \). 
Step 2: The equation becomes: \[ \frac{1}{2} \times 5 \times 10^{-3} \times (10)^2 = 5 \times 10^{-6} \times V \] \[ \Rightarrow 0.25 = 5 \times 10^{-6} \times V \] 
Step 3: Solving for \( V \): \[ V = \frac{0.25}{5 \times 10^{-6}} = 50 \, {kV} \] 
Thus, the required potential difference is 50 kV.