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

Moderate

Solution

Net force on the particle should be zero.
$∴ E q = 200 \times 10^{- 6} N = 2 \times 10^{- 4} N$
$m g = 2 \times 10^{- 5} \times 9.8$ $= 1.96 \times 10^{- 4} N$
Since $E q > m g$ , force $q v B$ should act downward to balance the forces.
$∴ E q = m g + q v B \Rightarrow 2 \times 10^{- 4} = 1.96 \times 10^{- 4} + (10^{- 6} \times v \times 2.0) \Rightarrow v = \frac{0.04 \times 10^{- 4}}{10^{- 6} \times 2.0} = 2 m / s$
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