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Early Days- Democritus and Dalton

In the assemb...

Question

In the assembly as shown below, the potential differences across the plated is 4 volts. A positive particle of charge +4e is projected from the negative plate with an initial kinetic energy of 4eV and the negative particle of charge (-2e) is projected from the positive plate.Both the particle reach point A with zero kinetic energy. Find the initial energy of the negative particle in eV.

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Solution

For the positive particle, applying energy conservation initially and at a point A

$K . E_{i} + P . E_{i} = K . E_{f} + P . E_{f}$

$⟹ 4 e V + (+ 4 e) (0 V) = 0 + (+ 4 e) (x v o l t)$ {X=potential at point A}

$⟹ x = 1 v o l t$

Now applying energy conservation for the negative particle at point A and initially

$⟹ K . E_{i} + (- 2 e V) (4 V) = 0 + (- 2 e) (1 v o l t)$

$K . E_{i} - 8 e V = - 2 e V$

$K . E_{i} = 6 e V$

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