Question

There are two large conducting plates kept parallel to each other It is found that charge exist only on the inner faces of plates facing each other the magnitude of surface charge density on the metal face is Sigma the separation between the two plates is 1 millimetre and an electron starting from rest near the one plate takes 1 nanosecond to strike The Other plate what will be the magnitude of Sigma

Answer 1

Given:

The two large conducting plates are kept parallel to each other .

Charge is present only on inner faces of the plates facing each other .

Magnitude of surface charge density on metal face = $\sigma$

Distance between the two conducting  plates = 1 mm

Time taken by an electron which starts from near one of the plate to strike the other plate = 1 nano sec

To Find :

Magnitude of $\sigma$ = ?

Solution :

Since we know that electric field at a point P between two conducting plates kept parallel to each other = $\frac{\sigma }{\epsilon_0}$

Now the force F acting on the electron :

$F = q_e \times E$

Her , $q_e$ is charge on electron and E is electric field .

So, $F = q_e\frac{\sigma}{\epsilon_0}$    -(1)

Also , $F = m_e \times a$

Here , $m_e$ is mass of electron and a is acceleration of electron .

So, $a = \frac{F}{m_e}$

Putting  value of F from eq 1 in above equation , we have  :

$a = \frac{q_e . \frac{\sigma}{\epsilon_0}}{m_e} = \frac{q_e \sigma }{m_e \epsilon_0}$

Now from 2nd equation we have :

$S = ut + \frac{1}{2} at^2$

Putting all the values in this equation we have ;

$1 \times 10^{-3} = \frac{1}{2} \times (\frac{q_e . \sigma }{m_e . \epsilon_0})\times t^2$

Or, $\sigma = \frac{2 \times 9.1 \times 8.85 \times 10^{-31} \times 10^{-12} }{1.6 \times 10^{-19} \times 10^{-18}}$

Or, $\sigma = 1.0066 \times 10^{-4} \frac{C}{m^2}$

So, the required value of sigma is $\sigma = 1.0066 \times 10^{-4} \frac{C}{m^2}$