Question

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?

Answer 1

Answer:

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 = $\frac{1.6\times10^{-19} \times1 \times 10^{-9} \times2\times10^{-3} }{2\times4\times10^{-4}\times 8.85 \times10^{-12} }$

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 = $\frac{2\times4.52\times10^{-17} }{9.1\times10^{-31} }$

v = $\sqrt{0.99\times10^{14} }$

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 = $\frac{2\times4.52\times10^{-17} }{1.67\times10^{-27} }$

v = $\sqrt{5.41\times10^{10} }$

v = 2.32 x 10⁵ m/s

Answer 2

(a) The energy of each particle is 9 × 10⁻¹⁷ J

(b) Speed of electron is 1.4 × 10⁷ m/s

The speed of the proton is 3.35 × 10⁵ m/s

Potential difference and energy:

Given that the parallel plates have dimensions 2.0 cm * 2.0 cm, their area will be:

A = 2×10⁻² × 2×10⁻²

A = 4×10⁻⁴ m²

The distance between the plates is d = 2mm = 2×10⁻³ m

The charge on the plates is Q = 1nC = 1×10⁻⁹ C

The relation between capacitance, voltage, and charge is:

Q = C x V

Now, capacitance:

C = (∈₀A)/d

So,

V = Q/C

V = Qd/(∈₀A)

Proton and electron both have an equal magnitude of charge, so both of them will have energy:

U = qV

where q is the charge

U = qQd/(∈₀A)

$U=\frac{1.6\times10^{-19}\times1\times10^{-9}\times2\times10^{-3}}{4\times10^{-4}\times8.85\times10^{-12}}$

U = 9 × 10⁻¹⁷ J

(b) Initially the particles have no kinetic energy only potential energy. When they arrive at the plate, their potential energy becomes zero.

According to the law of conservation of energy, all the potential energy is converted into kinetic energy:

KE = 9 × 10⁻¹⁷ J

¹/₂mv² = 9 × 10⁻¹⁷ J

v = $\sqrt{\frac{18\times10^{-17}}{m} }$

For electron:

v = $\sqrt{\frac{18\times10^{-17}}{9.1\times 10^{-31}} }$

v = 1.4 × 10⁷ m/s

For proton:

v = $\sqrt{\frac{18\times10^{-17}}{1.6\times 10^{-27}} }$

v = 3.35 × 10⁵ m/s

Learn more about conservation of energy:

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