Question

The electric field between two parallel plates is uniform, with magnitude 576 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 4.06 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. Ignore the electrical attraction between the proton and electron.

_______cm

(b)

Repeat part (a) for a sodium ion (Na⁺) and a chloride ion (Cl⁻). (Give your answer in cm.)

________cm

Answer 1

In the question the electric field between the two plates is given as 576 N/C. The plates are separated by a distance of 4.06 cm.

A proton is held stationary at the negative plate, while a proton is held stationary at the positive plate. When these are released, both the proton and electron will accelerated towards the oppositelly charged plate. It is mentioned in the question that we can ignore the force of attraction between the proton and the electron.

So, suppose the positively charged plate is on the right side and the negatively charged plate is on the left side. The electron in the negative plate will be accelerated towards the positive plate (from left to right), while the proton in the positive plate will be accelerated towards the negative plate (from right to left).

The magnitude of charge of the proton and the electron are same. We can denote it as 'q'.

So, the force acting on the proton or electron when it is released is given by,

... Equation (1)

Since the mass of the electron and the proton is different. The acceleration will vary for the proton and the electron.

Let be the mass of the proton and electron respectively.

So, the acceleration on the electron produced by the electric field.

... equation (2)

So, the acceleration on the proton produced by the electric field.

... equation (3)

Both the electron starts it motion from rest. So, the initial velocity of both the proton and electron can be taken as zero.

Consider a point 'x' from the negatively charged plate where the electron and proton pass each other at a partcular time 't', we can denote the rest of the distance as (0.0406m-x) m.

So, the equation of a proton and an electron can be written using the equation,

In our case the initial velocity u is zero in both cases. So, u=0.

In the case of an electron the distance s=x and a=a_e. So, we can write,

... equation (4)

In the case of an electron the distance s=x and a=a_p. So, we can write,

... equation (5)

Taking the ratio of equation (5) to equation (4), we get,

..... Equation (6)

The mass of the electron is and the mass of the proton is

So, the electron and the proton will cross each other at a distance of (4.06-4.057) cm=0.003 cm from the positive plate.

b) In the case of a sodium ion and a chlorine ion, the position of the proton in the above problem is filled by the sodium ion and the position of the electron is filled by the chlorine ion. The sodium ion has excess positive charge whose magnitude equal to the charge of proton and for the chlorine ion it has excess negative charge, whose magnitude is equal to the charge of electron.

So, the force acting on these chlorine ion and sodium ion will be same.

So, the steps for the solution is same as given in the problem above. In the case of electron and proton mass we have to substitute chlorine ion mass and sodium ion mass.

So, using equation (6), we can write,

Mass of the sodium ion is and the mass of the chlorine ion is . So, substituting these vaues in the above equation, we get,

So, the chlorine ion and the sodium ion will pass each other at a distance of 2.4 cm from the positive plate.