Question

If a particle of mass 10 mg and charge 2 µC at rest is subjected to a uniform electric field of potential difference 160 V, then the velocity acquired by the particle is

• A. 9 ms⁻¹
• B. 4 ms⁻¹
• C. 6 ms⁻¹
• D. 8 ms⁻¹

Hint:

Always convert all given quantities to their base SI units (mass to kg, charge to C, etc.) before substituting them into formulas. This prevents unit-related errors in the final calculation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem applies the Work-Energy Theorem. When a charged particle is accelerated by an electric field through a potential difference, the work done by the electric field on the particle is converted into its kinetic energy.
Step 2: Key Formula or Approach:
The work done (\(W\)) by the electric field on a charge (\(q\)) moving through a potential difference (\(\Delta V\)) is given by:
\[ W = q \Delta V \] The change in kinetic energy (\(\Delta K\)) of the particle is:
\[ \Delta K = K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \] According to the Work-Energy Theorem, \(W = \Delta K\).
Step 3: Detailed Explanation:
First, we list the given values in SI units:
- Mass, \(m = 10 \text{ mg} = 10 \times 10^{-3} \text{ g} = 10 \times 10^{-6} \text{ kg}\).
- Charge, \(q = 2 \text{ µC} = 2 \times 10^{-6} \text{ C}\).
- Potential difference, \(\Delta V = 160 \text{ V}\).
- The particle starts from rest, so its initial velocity, \(u = 0\).
Now, we apply the Work-Energy Theorem:
\[ q \Delta V = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2 \] \[ q \Delta V = \frac{1}{2}mv^2 \] We rearrange the formula to solve for the final velocity, \(v\):
\[ v^2 = \frac{2q \Delta V}{m} \] Substitute the given values:
\[ v^2 = \frac{2 \times (2 \times 10^{-6} \text{ C}) \times (160 \text{ V})}{10 \times 10^{-6} \text{ kg}} \] The \(10^{-6}\) terms in the numerator and denominator cancel out.
\[ v^2 = \frac{2 \times 2 \times 160}{10} = \frac{640}{10} = 64 \] Now, take the square root to find the velocity:
\[ v = \sqrt{64} = 8 \text{ m/s} \] Step 4: Final Answer:
The velocity acquired by the particle is 8 ms⁻¹. Therefore, option (D) is correct.