The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 imes 103 \mathrm{Vm}^{-1}$ acting vertically upwards as shown in the figure. The plates are sufficiently long and have separation $2 \mathrm{~cm}$. A particle of negative charge $1 \mu \mathrm{C}$ and…

Question

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 103 \mathrm{Vm}^{-1}$ acting vertically upwards as shown in the figure.

The plates are sufficiently long and have separation $2 \mathrm{~cm}$. A particle of negative charge $1 \mu \mathrm{C}$ and mass $2 \mathrm{~g}$ is projected at an angle $45^{\circ}$ with the electric field from the lower plate with a velocity ' $u$ '. The maximum velocity acquired by the particle, if it is not hit the upper plate is

Aicharya · Accepted Answer

We have, $ \begin{aligned} & h_{\max }=\frac{u^2 \sin ^2 \theta}{2\left(\frac{q E}{m}\right)}\left[\text { Here } g \rightarrow \frac{q E}{m}\right] \\ & u^2=\frac{2 h m q E}{m \sin ^2 45^{\circ}} \quad u^2=\frac{2 \times 2 \times 10^{-2} \times 10^{-6} \times 10^3}{2 \times 10^{-3} \times 1 / 2} \\ & \Rightarrow u^2=0.04 \Rightarrow u=0.2 \mathrm{~m} / \mathrm{s} \end{aligned} $

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 103…

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