Question

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.22 cm. At the same moment, both particles are released.
A. Calculate the distance (in cm) from the positive plate at which the two pass each other.
B. Repeat part (a) for a sodlum lon (Nat) and a chlorlde lon (CI).

Answer 1

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$E

and, we know that, F = ma

So,

$m_{p}$a = $q_{p}$E

a = $\frac{q_{p}.E }{m_{p} }$      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2$at^{2}$

Here, u = 0.

S = 1/2$at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ($\frac{q_{p}.E }{m_{p} }$  )$t^{2}$      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ($\frac{q_{e}.E }{m_{e} }$  )$t^{2}$    Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S$m_{p}$ = $m_{e}$D - $m_{e}$S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm