# 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 - Physics

Sum

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?

#### Solution

Let the electric field inside the capacitor be E.

Now ,

Magnitude of acceleration of the electron, a_e = (q_eE)/m_e

Magnitude of acceleration of the proton, a_p = (q_pE)/m_p

Let t be the time taken by the electron and proton to reach the positive and negative plates, respectively.
The initial velocities of the proton and electron are zero.
Thus, the distance travelled by the proton is given by

x = 1/2 (q_pE)/(m_p)t^2                .....(1)

And, the distance travelled by the electron is given by

2-x = 1/2 (q_eE)/(m_e)t^2            ......(2)

On dividing (1) by (2), we get

x/(2-x) = ((qpE)/(mp))/((qeE)/(m_e)) = m_e/m_p = (9.1 xx 10^-31)/(1.67 xx 10^-27) = 5.449 xx 10^-4

⇒ x = 10.898 xx 10^-4 - 5.449  xx 10^-4 x

⇒ x = (10.898 xx 10^-4)/1.0005449 = 1.08 xx 10^-8  "cm"

Concept: Combination of Capacitors
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 23 | Page 167