Question

The plates of a capacitor are 2⋅00 cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released?

Answer 1

The pair was released at the distance of 1.08×10-8 cm.

Explanation:

Inside the capacitor, the electric field is E.

Magnitude of electron’s acceleration, ae = qeEme

Magnitude of proton’s acceleration, ap = qpEmp

The time taken by the proton and electron to reach the negative and positive plates, respectively is t.

Zero is the initial velocities of the electron and proton.

So, the distance moved by the proton is  

X = 12qpEmpt2          … (1)

And, the distance moved by the electron is  

2-x = 12qeEmet2       …(2)

On (1) is divided by (2), we get

$\Rightarrow x=10.898 \times 10-41.0005449=1.08 \times 10-8 \mathrm{cm}$

Answer 2

Answer:

Explanation:

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