WBJEE · Physics · Electrostatics
A particle with charge \(e\) and mass \(m\), moving along the \(X\) -axis with a uniform speed \(u\), enters a region where a uniform electric field \(E\) is acting along the \(Y\) -axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is
- A \(\frac{2 m u^{2}}{e E}\)
- B \(\frac{e E}{2 m u^{2}}\)
- C \(\frac{m u}{2 e E}\)
- D \(\frac{m u^{2}}{2 e E}\)
Answer & Solution
Correct Answer
(D) \(\frac{m u^{2}}{2 e E}\)
Step-by-step Solution
Detailed explanation
From parabola \(y=\frac{1}{2} \times\left[\frac{E e}{m}\right] \times \frac{x^{2}}{u^{2}}\) \(=\frac{E \cdot e}{2 m u^{2}} \cdot X^{2}\) \(\begin{array}{ll}\text { As } x=4 a y & \text { (for parabola) }\end{array}\) \(\therefore \quad X^{2}=\frac{2 m u^{2}}{E \cdot e} \cdot y\)…
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