JEE Mains · Physics · STD 12 - 1. Electric charges and fields
An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity \(10^6 \mathrm{~m} / \mathrm{s}\). If the magnitude of the electric field between the plates is \(9.1 \mathrm{~V} / \mathrm{cm}\), then the vertical component of velocity of electron is (mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}\) and charge of electron \(=1.6 \times 10^{-19} \mathrm{C}\) )
- A 0
- B \(1 \times 10^6 \mathrm{~m} / \mathrm{s}\)
- C \(16 \times 10^6 \mathrm{~m} / \mathrm{s}\)
- D \(16 \times 10^4 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(16 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \Rightarrow \mathrm{t}=\frac{\ell}{\mathrm{V}_{\mathrm{x}}}=\frac{10 \times 10^{-2}}{10^6}=10^{-7} \\ & \Rightarrow \mathrm{~V}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t} \\ & \Rightarrow…
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