Question

A particle of mass $2\times 10^{- 5} \, kg$ moves horizontally between the plates of a parallel plate capacitor which produce an electric field of $200 \, N \, C^{- 1}$ in the vertically upward direction. A magnetic induction of $\text{2.0} \, T$ is applied at right angles to the electric field in a direction normal to both $\overset{ \rightarrow }{\text{E}}$ and $\overset{ \rightarrow }{\text{v}}$ . If $g$ is $\text{9.8} \, m \, s^{- 2}$ and the charge on the particle is $10^{- 6} \, C$ , then the velocity of the charged particle so that it continues to move horizontally is

• A. $2ms^{- 1}$
• B. $20ms^{- 1}$
• C. $0.2ms^{- 1}$
• D. $100ms^{- 1}$

Answer 1

Answer: (A)

Net force on the particle should be zero.



$\text{qE} = 1 0^{- 6} \times 2 0 0 = 2 \times 1 0^{- 4} N$
$\text{mg} = 2 \times 1 0^{- 5} \times \text{9.8} = \text{1.96} \times 1 0^{- 4} N$
Since $qE>mg$ , so magnetic force $qvB$ should act downwards to balance the forces.
$\text{qE} = \text{mg} + \text{qvB} \Rightarrow 2 \times 1 0^{- 4} = \text{1.96} \times 1 0^{- 4} + 1 0^{- 6} \text{v} \times 2$
$\Rightarrow \text{v} = 2 m/s$