KCET · Physics · Electrostatics
An electron of mass \( \mathrm{m} \), charge e falls through a distance \( \mathrm{h} \) meter in a uniform electric field \( \mathrm{E} \). Then time of fall
- A \( t=\sqrt{\frac{2 h m}{e E}} \)
- B \( t=\frac{2 h m}{e E} \)
- C \( t=\sqrt{\frac{2 e E}{h m}} \)
- D \( t=\frac{2 e E}{h m} \)
Answer & Solution
Correct Answer
(A) \( t=\sqrt{\frac{2 h m}{e E}} \)
Step-by-step Solution
Detailed explanation
Given, mass of electron \( =m \); charge \( =e \); height \( =h \)
Using \( S+u t+\frac{1}{2} a t^{2} \), we have
\(h=0+\frac{1}{2} \frac{e E}{m} t^{2}\)
\(\Rightarrow h=\frac{1}{2} \frac{e E}{m} t^{2}\)
On rearranging, we get
\(t^{2}=\frac{2 m h}{e E} \)
\(\Rightarrow t=\sqrt{\frac{2 m h}{e E}}\)
Using \( S+u t+\frac{1}{2} a t^{2} \), we have
\(h=0+\frac{1}{2} \frac{e E}{m} t^{2}\)
\(\Rightarrow h=\frac{1}{2} \frac{e E}{m} t^{2}\)
On rearranging, we get
\(t^{2}=\frac{2 m h}{e E} \)
\(\Rightarrow t=\sqrt{\frac{2 m h}{e E}}\)
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