When a positively charged particle is projected with a velocity v 0 in the region between the horizontal plates of the capacitor then it leaves by making an angle θ with the horizontal, as shown. If θ

Question

When a positively charged particle is projected with a velocity $v_{0}$ in the region between the horizontal plates of the capacitor then it leaves by making an angle $θ$ with the horizontal, as shown. If $θ = θ_{1}$ is for proton and $θ = θ_{2}$ for an $α$ -particle then the value of $\frac{\tan θ_{1}}{\tan θ_{2}}$ is

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Answer:

C

Solution:

$\tan θ = \frac{V_{y}}{V_{x}}$

$\Rightarrow \tan θ \propto \frac{q}{m}$

$\Rightarrow \frac{\tan θ_{1}}{\tan θ_{2}} = 2$

TestHub · Accepted Answer

tan ⁡ θ = V y V x ⇒ tan ⁡ θ ∝ q m ⇒ tan ⁡ θ 1 tan ⁡ θ 2 = 2

Physics - Electrostatics Question with Solution | TestHub

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