Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What Do You Mean by the Electrical Potential Energy of a System of Charges?
The electrical potential energy of a system of charges is equal to the total work done by an external agent in order to bring the charges, one by one, from infinite separation to the
desired positions to form the system.
Let q1 and q2 be two charges placed at the points A and B
respectively and AB = r [Fig.]. To determine the potential energy of the system of the two charges q1 and q2, let us suppose that in the region only the charge q1 existed initially but q2 was absent.
The potential at B due to the charge q1 is,
V = \(\frac{1}{4 \pi \epsilon_0} \frac{q_1}{r}\)
Now to bring the charge q2 from infinity to the point B, work done,
W = Vq2
This work done gets stored in the system of charges q1 and q2, and is called the potential energy of the system (U).
So, U = Vq2 = \(\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}\) …. (1)
In CGS system, ε0 is to be replaced by \(\frac{1}{4 \pi}\) in equation (1). The units of U in SI is joule (J) and in CGS is erg.
To determine the potential energy of a system of multiple charges, the above procedure has to be repeated step by step.
For a system of charges q1, q2, q3, q4 ………, we have to continue the calculations after equation (1) as follows:
- work done to bring the charge q3 from infinity to a point near the already formed system of q1 and q2;
- work done to bring the charge q4 from infinity to a point near the already formed system of q1, q2 and q3; and so on.
The algebraic sum of all the above values of work done would then be equal to the potential energy stored in the system of charges.
Clearly, the procedure is often extremely complicated. However, the calculations turn out to be fairly easy for some systems with specific symmetries.
Calculation of electrostatic potential energy for a system of three point charges:
The potential energy of a system of threé charges q1, q2 and q3 is equal to the total work done to bring these charges one by one from infinity to the positions and respectively.
Calculation of electrostatic potential energy for a system of three point charges:
The potential energy of a system of threé charges q1, q2 and q3 is equal to the total work done to bring these charges one by one from infinity to the positions and respectively.
To bring the charge q1 at the position \(\vec{r}_1\), no work is done, as all other charges are still at infinity, i.e., there is no field [Fig.].
So, W1 = 0
To bring the charge q2 from infinity to the position at a distance r12 from q1, work is done
W2 = [potential for q1] × q2] = \(\frac{q_1}{4 \pi \epsilon_0 \kappa r_{12}}\) = \(\) × q2 = \(\frac{q_1 q_2}{4 \pi \epsilon_0 k r_{12}}\)
Similarly, to bring the charge q3 from infinity to the position \(\overrightarrow{r_3}\), work has to be done against electrostatic forces of both q1 and q2,
Work Done in Deflecting a Dipole in a Uniform Electric Field and Potential Energy of the Dipole
Consider an electric dipole placed at an angle θ with a uniform electric field of strength E. The dipole moment of the electric dipole is p. We know that the torque acting on it is,
\(\tau\) = pEsinθ
This torque acting on the dipole brings it along the direction of the electric field. Obviously when θ = 0°, the dipole is in equilibrium position. To deflect It from its equilibrium position
work has to be done on the dipole. This work is stored up as potential energy in the dipole at Its deflected position.
Suppose, the dipole is in equilibrium position. An external torque acts on it and deflects it through an angle θ [Fig.]. In this position the magnitude of the torque applied by the external agent is,
\(\tau_{\text {ext }}\) = PEsinθ
So, work done to deflect the dipole from its equilibrium position to an angle θ is
Therefore, in this position potential energy of the electric dipole is given by,
U = pE(1 – cosθ) …….. (2)
From the equation (1) we can say that to deflect the dipole
- from the equilibrium position through an angle 90°, work done, W = pE
- from the equilibrium position through an angle 180°, work done, W = 2pE
- from angle θ to angle θ2,
work done, W = pE( cosθ1 – cosθ2)
Again, let the dipole at first be at right angles to the electric field. Then to bring the dipole from that position to an angle θ, the change in potential energy is given by,
The work done on a dipole is equal to the difference in potential energies between its two orientations. As the initial value of potential energy is not physically significant, so it can be taken as zero for any standard orientation of this dipole.
If we take potential energy of the dipole to be zero when it remains at right angles to the electric field, then U\(\left(\frac{\pi}{2}\right)\) = 0
Therefore in that case,
U(θ) = –\(\vec{p} \cdot \vec{E}\) ……… (3)
Kinetic Energy of a Charged Body in an Electric Field
We know that, when a body falls freely from a height under the action of gravity, it loses its potential energy but gains an equal amount of kinetic energy. Similarly when a charged body moves freely in an electric field, it loses a certain amount of potential energy and gains an equal amount of kinetic energy.
Suppose a particle of charge q is moving from one point to another in an electric field. If the potential difference between the two points is V, the particle loses potential energy of amount qV. So the increase of its kinetic energy will also be,
Ek = qV …. (1)
If m be the mass of the particle and v1, v2 be its velocities at the first and the second points, respectively, then
Ek = \(\frac{1}{2} m\left(v_2^2-v_1^2\right)\) …. (2)
∴ \(\frac{1}{2} m\left(v_2^2-v_1^2\right)\) = qV ….. (3)
If the particle starts from rest and acquires a velocity v at the second point then,
\(\frac{1}{2} m v^2\) = qV or, v = \(\sqrt{\frac{2 q V}{m}}\) …….. (4)
With the help of this equation, velocities of charged particles like electron, proton, etc., are determined in different atomic and nuclear experiments.
Electronvolt: it is a special unit of energy. The energies of particles like electron, proton, etc., in atomic and nuclear physics are measured in this unit.
Definition: The amount of kinetic energy acquired by an electron, when it accelerates through a potential difference of 1 volt, is called an electronvolt (eV).
1 electronvolt (1 eV)
= charge of an electron × 1V
= 4.8 × 10-10 esu of charge × \(\frac{1}{300}\) esu of potential
= 1.6 × 10-12 erg
= 1.6 × 10-19J [∵ 1 erg = 10-7 J]
Bigger units like kiloelectronvoit (keV), megaelectronvolt(MeV), etc. are also used.
1keV = 103 eV = 1.6 × 10-9 erg = 1.6 × 10-16 J
1MeV = 106 eV = 1.6 × 10-6 erg = 1.6 × 10-13 J
Numerical Examples
Example 1.
Two point charges of +49 esu and +81 esu are placed at a separation of 100 cm in air. Determine the position of the neutral point in the electric fields of the two charges. What is the electric potential at the neutral point?
Solution:
The neutral point must be in between the two charges because both the charges are positive. Let +49 esu and +81 esu of charges he placed respectively at A and B [Fig.]
Let the neutral point P he at a distance x cm from the point A. So the distance of P from the point B is (100 – x) cm.
According to the question,
\(\frac{49}{x^2}\) = \(\frac{81}{(100-x)^2}\) or, \(\frac{7}{x}\) = \(\frac{9}{100-x}\) or, x = 43.75 cm
So the neutral point is on the line AB at a distance of 43.75 cm from the charge +49 esu.
If V be the potential at the neutral point, then
V = \(\frac{49}{43.75}\) + \(\frac{81}{56.25}\) = 1.12 + 1.44 = 256 esu of potential
Example 2.
An electron is subjected to a potential difference of 180 V. Mass and charge of an electron are 9 × 10-31 kg and 1.6 × 10-19 C, respectively. Find the velocity acquired by the electron.
Solution:
Here, mass of the electron, m = 9 × 10-31 kg
Charge of the electron, e = 1.6 × 10-19 C
Potential difference, V = 180 V
∴ Velocity acquired by the electron, v is given by
\(\frac{1}{2} m v^2\) = eV
or, v = \(\sqrt{\frac{2 e V}{m}}\) = \(\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 180}{9 \times 10^{-31}}}\) = 8 × 106 m ᐧ s-1
Example 3.
At each of the four vertices of a square of side 10 cm, four positive charges each of 20 esu are placed. Find the potential at the point of intersection of the two diagonals.
Solution:
Amount of charge at each corner, q = 20 esu of charge
Length of each side of the square = 10 cm
So, length of the diagonal = \(\sqrt{10^2+10^2}\) = 10\(\sqrt{2}\) cm
∴ The distance of a vertex from the point of intersection of the diagonals is,
∴ x = \(\frac{1}{2}\) × length of the diagonal = \(\frac{1}{2}\) × \(10 \sqrt{2}\) = \(5 \sqrt{2}\)
∴ Potential at the point of intersection of the two diagonals
= \(\frac{20}{5 \sqrt{2}}\) + \(\frac{20}{5 \sqrt{2}}\) + \(\frac{20}{5 \sqrt{2}}\) + \(\frac{20}{5 \sqrt{2}}\)
= \(\frac{80}{5 \sqrt{2}}\) = 8\(\sqrt{2}\) = 11.31 stat V
Example 4.
The distance between two points A and B in vacuum is 2d. At each of these two points a +Q charge is placed. P is the midpoint of AB. Find the intensity and potential at P due to the electric field. How will the values of these quantities change if the charge at B is replaced by a charge -Q ?
Solution:
AB = 2d; mid point of AB is P, i.e., AP = BP = d [Fig.]
Intensity at P due to the charge + Q at A
= \(\frac{Q}{d^2}\); along \(\overrightarrow{P B}\) ….. (1)
Intensity at P due to the charge + Q at B
= \(\frac{Q}{d^2}\); along \(\overrightarrow{P A}\) ….. (2)
So, intensities at P due to the charges at A and B are equal and opposite.
Therefore, the resultant intensity at P is zero.
Potential at P due to the charge at A = \(\frac{Q}{d}\)……. (3)
Potential at P due to the charge at B = \(\frac{Q}{d}\)…….. (4)
∴ Total potential at P = \(\frac{Q}{d}\) + \(\frac{Q}{d}\) = \(\frac{2 Q}{d}\)
Now if -Q charge be placed at the point B, intensity in equation (2) will be \(\frac{Q}{d^2}\) along \(\overrightarrow{P B}\).
∴ Resultant intensity at P = \(\frac{Q}{d^2}\) + \(\frac{Q}{d^2}\) = \(\frac{2 Q}{d^2}\); along \(\overrightarrow{P B}\)
Again the value of potential in equation (4) will be \(\frac{-Q}{d}\).
∴ Total potential at P = \(\frac{Q}{d}\) – \(\frac{Q}{d}\) = 0
Example 5.
Four point charges of +100 esu, -50 esu, +20 esu and +30 esu are placed respectively at the four vertices A, B, C, D of a square of side 10 cm. Find the electric intensity and potential at the point of intersection of the diagonals. [HS 2000]
Solution:
Length of the diagonal of the square
= \(\sqrt{10^2+10^2}\) = 10\(\sqrt{2}\) cm
The point of intersection of the two diagonals is the point O [Fig.].
Intensity acts along \(\overrightarrow{O C}\) due to the charge at A and it acts along \(\overrightarrow{O A}\) due to the charge at C.
So resultant intensity at O due to the charges at A and C is given by
E1 = \(\frac{100}{(5 \sqrt{2})^2}\) – \(\frac{20}{(5 \sqrt{2})^2}\)
= (2 – \(\frac{2}{5}\)) = \(\frac{8}{5}\) dyn statC-1; along \(\overrightarrow{O C}\)
Similarly, intensity at O due to the charges at B and D acts along \(\overrightarrow{O B}\).
So, resultant intensity at O due to the charges at B and D is given by,
E2 = \(\frac{30}{(5 \sqrt{2})^2}\) + \(\frac{50}{(5 \sqrt{2})^2}\)
= (\(\frac{3}{5}\) + 1) = \(\frac{8}{5}\) dyn ᐧ statC-1, along \(\overrightarrow{O B}\)
Here, E1 and E2 act perpendicular to each other and E1 = E2.
∴ Resultant intensity at O,
E = \(\sqrt{E_1^2+E_2^2}\) = \(\sqrt{\left(\frac{8}{5}\right)^2+\left(\frac{8}{5}\right)^2}\) = \(\frac{8}{5} \sqrt{2}\)
= 2.263 dyn statC-1
The direction of E is along the bisector \(\overrightarrow{O P}\) of the angle COB, i.e., parallel to the side AB or DC.
Again, potential at O = \(\frac{100}{5 \sqrt{2}}\) – \(\frac{50}{5 \sqrt{2}}\) + \(\frac{20}{5 \sqrt{2}}\) + \(\frac{30}{5 \sqrt{2}}\)
= \(\frac{100}{5 \sqrt{2}}\) = 10\(\sqrt{2}\) = 14.14 statV
Example 6.
Electrons starting from rest and passing through a potential difference of 60 kV are found to acquire a velocity of 1.46 × 1010 cm ᐧ s-1. Calculate the ratio of
charge to the mass of an electron.
Solution:
Here, velocity of an electron,
v = 1.46 × 1010 cm ᐧ s-1 = 1.46 × 108 m ᐧ s-1
Potential difference, V = 60 kV = 60000 V
If e be the charge and m be the mass of an electron and v be the velocity acquired by it in passing through a potential difference V, we have,
\(\frac{1}{2} m v^2\) = eV
or, \(\frac{e}{m}\) = \(\frac{v^2}{2 V}\) = \(\frac{\left(1.46 \times 10^8\right)}{2 \times 60000}\) = 1.776 × 1011 m ᐧ kg-1
Example 7.
A particle charged with 1.6 × 10-19 C is in motion. It enters the space between two parallel metal plates, parallely along the midway between them. The plates are 10 cm long and the separation between them is 2 cm. A potential difference of 300 V exists between the plates. Find out the maximum velocity of the charged particle at the point of entry, for which it would be unable to emerge from the space between the plates. Given, mass of the particle = 12 × 10-24 kg.
Solution:
Length of the plates A and B [Fig.] is l = 10 cm = 0.1 m
Half of the distance between the plates,
d = \(\frac{1}{2}\) × 2 cm = 1 cm = 0.01 m
Uniform electric field in the intermediate space,
E = \(\frac{300 \mathrm{~V}}{2 \mathrm{~cm}}\) = \(\frac{300 \mathrm{~V}}{0.02 \mathrm{~m}}\) = 15 × 103 V ᐧ m-1
Let the charge q = 1.6 × 10-19 be positive.
So, a downward force acts on it due to the electric field E, given by
F = qE
∴ Downward acceleration, a = \(\frac{q E}{m}\)
At the point of entry, let u be the velocity of the charged particle.
In the axial direction, it experiences no force; so the time taken to travel the distance l is t = \(\frac{l}{v}\)
Again, at the point of entry, the downward component of velocity = 0; Thus the downward displacement in time t is
x = \(\frac{1}{2} a t^2\) = \(\frac{1}{2} \frac{q E}{m}\left(\frac{l}{v}\right)^2\)
The condition, that the particle will not emerge from the space between the plates, is
Example 8.
An infinite number of charges, each of value q, are placed on the x -axis at the points x = 1, x = 2, x = 4,x = 8, …….. Find the potential and intensity due to these charges at x = 0. If the charges are alternately positive and negative what will be the potential and intensity at the same point?
Solution:
If V be the potential at x = 0, then
In the second case, when the consecutive charges are of opposite signs, let us assume that the first charge is positive. Then potential,
Example 9.
Two equally charged soap bubbles of equal volume are Joined together to form a larger bubble. If each bubble had a potential V, find the potential of the resultant bubble. [WBJEE 2000]
Solution:
Let the radius of the smaller bubble be r and that of the larger bubble be R. Charge of each bubble = q.
According to the question,
\(\frac{4}{3} \pi R^3\) = \(2 \cdot \frac{4}{3} \pi r^3\) or, R = 21/3 ᐧ r
Potential of each soap bubble,
V = \(\frac{q}{r}\) or, q = Vr
∴ Potential of the larger bubble
\(=\frac{\text { total charge }}{\text { radius }}\) = \(\frac{2 q}{R}\) = \(\frac{2 V r}{R}\) = \(\frac{2 V r}{2^{1 / 3} r}\) = 22/3 V
Example 10.
An electric dipole of moment 5 × 10-8 C ᐧ m is placed in an electric field of magnitude 4 × 105 N ᐧ C-1. What amount of work is to be done to deflect it through an angle 60°?
Solution:
We know, work done,
W = pE(1 – cosθ)
= 5 × 10-8 × 4 × 105(1 – cos 60°)
Here, p = 5 × 10-8 C ᐧ m, E = 4 × 105N × C-1, θ = 60°
∴ W = 2 × 10-2(1 – 0.5)
= 10-2 J