Question

The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is The capacitor is cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

Answer 1

Answer:

The electric field between the plates is 2173 N/C

Explanation:

Given that,

Distance = 0.150 cm

Suppose,The initial speed of the electron is $7.05\times10^6\ m/s$. The capacitor is 2.00 cm long,

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{2.00\times10^{-2}}{7.05\times10^{6}}$

$t=2.8\times10^{-9}\ s$

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

Put the value into the formula

$a=\dfrac{2\times0.150\times10^{-2}}{(2.8\times10^{-9})^2}$

$a=3.82\times10^{14}\ m/s^2$

We need to calculate the electric field between the plates

Using formula of electric field

$E=\dfrac{F}{q}$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{9.1\times10^{-31}\times3.82\times10^{14}}{1.6\times10^{-19}}$

$E=2173\ N/C$

Hence, The electric field between the plates is 2173 N/C