Question

1. A positive ion of charge 1 uC with a weight of 2 x 10-6 kg is released from the positive plate of a capacitor. When it reaches the negative plate, it has achieved a velocity of v = 100 m/s. Calculate the following: (6 marks total) a. What is the potential difference between the two plates? (2 marks) b. If the distance between the two plates is 1 nm and filled with vacuum, and the dimensions of the plates are 10cm by 20cm, calculate the capacitance of the capacitor. (2 marks) c. Calculate the electrical potential energy stored in the capacitor. (2 marks)

Answer 1

Given

$q=1\mu C$

$m=2*10^{-6}kg$

$v=100m/s$

a)

Initially the positive ion was at reston positive plate of capacitor.

It achieves velocity "v" till it reaches the negative plate.

The total change in kinetic energy is given by

ΔKE = EmΔυ-

Hence

KE final KEinitial KE - 0= = * 2 * 10-6 * (1002 – 02)

Thus $KE=0.01J$

This kinetic energy is the result of the stored potential energy in the positive ion during charging of the capacitor.

i.e. $KE=\Delta PE$

$KE=\frac{1}{2}q\Delta V$ $\Rightarrow 0.01=\frac{1}{2}* 1\mu *V$

Here $\Delta V$ Is the potential difference between the capacitor plates.

Let $\Delta V=V$ ...... Considering one of the plates is at 0V.

Thus V= 20,000V

........................................... .... . . . .....................................

b)

distance between the plates

Area of the plates

Permittivity $\epsilon=\epsilon_o \epsilon_r$

Where $\epsilon_o =$ absolute permittivity of free space =$8.854*10^{-12}$

$\epsilon_r =$ relative permittivity of a medium

$\epsilon_r = 1$ for vaccum

Capacitance of a parallel plate capacitor is given by

$C=\epsilon_o \epsilon_r \frac{A}{d}$

Substituting the values

We get

................................                 .............................................

C)

Potential energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

Substituting the values obtained above

We get $U=\frac{1}{2}*177.03*10^{-12}*20000^2= 17.7*10^{-3}J$

i e.

.

.

.

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