A 2.0 cm * 2.0 cm parallel-plate capacitor with a 2.0 mm spacing is charged to {1.0 nC. First a proton and then an electron are released from rest at the midpoint of the capacitor. a. What is each particle’s energy? b. What is each particle’s speed as it reaches the plate?

a. Energy of each particle is -4.52 x 10⁻¹⁷ J.

b. Speed of electron is 0.99 x 10⁷ m/s.

  Speed of proton is 2.32 x 10⁵ m/s .

Explanation:

a. Capacitance of parallel plate capacitor is given by:

C = (∈₀A)/2d

Here ∈₀ is permittivity of free space, A is area of capacitor and d is the spacing between the plates.

The relation between capacitance, voltage potential and charge is:

Q = C x V

Here Q is charge on the capacitor.

Potential of parallel plate capacitor is :

V = Q/C = Qd/(2∈₀A)

According to the problem, the final potential energy of proton and electron is zero.

So, there initial potential energy is :

U = qV

U = qQd/(2∈₀A)

Here q is charge of the particle.

Since the magnitude of charge of electron and proton are same, so there energy are also equal.

Put 1.6 x 10⁻¹⁹ C for q, 1 x 10⁻⁹ C for Q, 2 x 10⁻³ m for d, 4 x 10⁻⁴ m² for A and 8.85 x 10⁻¹² C²/N m² for ∈₀ in the above equation.

U =

U = 4.52 x 10⁻¹⁷ J

The energies of the particles are negative as the particles are attracted towards the plate. SO, there energy is -4.52 x 10⁻¹⁷ J.

b. As the particles are at rest, there initial kinetic energy is zero. Applying law of conservation of energy, the final kinetic energy of the particle is:

K.E. = U = 4.52 x 10⁻¹⁷ J      ....(1)

Since kinetic energy depends upon mass and velocity. So, the speed of electron and proton are different as there masses are different.

We first calculate the speed of electron with the help of equation (1):

0.5mv² = 4.52 x 10⁻¹⁷ J

v² =

v =

v = 0.99 x 10⁷ m/s

Now, we calculate the speed of proton with the help of equation (1):

0.5mv² = 4.52 x 10⁻¹⁷ J

v² =

v =

v = 2.32 x 10⁵ m/s

12000j is work done