Question

A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of \(6 \times 10^4\) N/C. Then the magnitude of the charge is

• A. \(5 \times 10^{-4}\) C
• B. \(5 \times 10^{-10}\) C
• C. \(-18 \times 10^{-6}\) C
• D. \(-5 \times 10^{-9}\) C

Hint:

For stationary particle, \(qE = mg\), direction of E determines sign.

Answer 1

The Correct Option is B

To solve this problem, we need to determine the magnitude of the charge that, when placed in an electric field, experiences a force that balances its weight. Let's break down the problem step-by-step:

1. The particle is held stationary, which means that the electric force (\(F_e\)) acting upwards needs to balance the gravitational force (\(F_g\)) acting downwards.
2. \(F_g = mg\), where \(m\) is the mass of the particle and \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
   • Given: \(m = 0.003 \, \text{g} = 0.003 \times 10^{-3} \, \text{kg}\) (since 1 g = \(10^{-3}\) kg).
   • Thus, \(F_g = 0.003 \times 10^{-3} \times 9.8 \, \text{N}\).
   • Calculating: \(F_g = 2.94 \times 10^{-5} \, \text{N}\).
3. The electric force is given by the formula: \(F_e = qE\), where \(q\) is the charge, and \(E\) is the electric field strength.
   • Given: \(E = 6 \times 10^4 \, \text{N/C}\).
4. Setting the gravitational force equal to the electric force for equilibrium, we have \(F_g = F_e\), which translates to:
5. \(q \times 6 \times 10^4 = 2.94 \times 10^{-5}\).
   • Solving for \(q\):

\(q = \frac{2.94 \times 10^{-5}}{6 \times 10^4}\).

• Calculating:

\(q = 4.9 \times 10^{-10} \, \text{C}\).

1. Rounding up to match the option, we can consider \(q \approx 5 \times 10^{-10} \, \text{C}\).

Conclusion: The magnitude of the charge is \(5 \times 10^{-10} \, \text{C}\), which corresponds to the given correct option. Other options are incorrect based on the calculations.