WBJEE · Physics · Electrostatics
A pair of parallel metal plates are kept with a separation 'd'. One plate is at a potential \(+\mathrm{V}\) and the other is at ground potential. A narrow beam of electrons enters the space between the plates with a velocity \(\mathrm{v}_{0}\) and in a direction parallel to the plates. What will be the angle of the beam with the plates after it travels an axial distance L?
- A \(\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}}\right)\)
- B \(\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)\)
- C \(\sin ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}}\right)\)
- D \(\cos ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)\)
Answer & Solution
Correct Answer
(B) \(\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)\)
Step-by-step Solution
Detailed explanation
Hint: \(t=\frac{L}{v_{0}}\) \(v_{y}=\frac{e}{m} \cdot \frac{v}{d} \times \frac{L}{v_{0}}=\frac{e v L}{m d v_{0}}\) \(\tan \theta=\frac{v_{y}}{v_{x}}=\frac{e v L}{m d v_{0} \cdot v_{0}}\) \(\Rightarrow \theta=\tan ^{-1}\left(\frac{e v L}{m d v_{0}^{2}}\right)\)
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