Question

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

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

Answer 1

The correct option is A $2m{s}^{-1}$

The electric force on the particle $F_E=qE={10}^{-6}\times 200=2\times {10}^{-4}$
Weight of the particle acts downward. $mg=2\times {10}^{-5}\times 9.8=1.96\times {10}^{-4}$
Since $F_E>mg$, $F_{mag}$ has to act downward to make the net force zero.
$F_{mag}=(2-1.96)\times {10}^{-4}=4\times {10}^{-6}$
$\therefore {10}^{-6}\times v\times 2=4\times {10}^{-6}$
$\Rightarrow v=2m/s$