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Forces between multiple charges can be shown by an example, such as when synthetic clothing is removed from our bodies, during dry weather, causing a spark or crackling sound.
- A charge is a property of an atom. An atom is known to be charged in case it possesses an unequal number of electrons and protons.
- An atom is positively charged in case it has a lesser number of electrons than protons.
- It is negatively charged if it has more electrons than protons.
- Electric charge can be defined as the amount of energy or electrons that pass from one body to another.
- It is also the amount of deficit or excess electrons in a body.
- A neutral body will have an equal number of positive and negative charges. The protons are present in the nucleus of the atom while the electrons move in orbits around the nucleus thus it is the presence or absence of electrons which decides the charge of the body. The electrons act as charge carriers.
Read Also: Difference between Cations and Anions
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Key Terms: Force of charges, Coulomb’s law, Electrostatic force, Static electricity, Like and Unlike Charges, Electrons, Polarity, Principle of Superposition, Surface Charge Density
Force between Multiple Charges
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When you rub a plastic comb or a plastic scale to your hair and then take them near to tiny pieces of paper, the plastic body starts acting as a magnet and starts attracting the paper pieces to itself.
A System of Three Charges
- This is because the rubbing of the plastic body makes it gain electrons from your hair thus possessing a more negative electric charge.
- These negative charges attract the positive charges already present in the tiny pieces of paper making them stick to the plastic body.
- The attraction or repulsion of differently charged bodies is due to the force of charges acting on them.
Magnitude of Force Between Two Charges
The force between any two charges can be determined by Coulomb’s law. Coulomb’s law claims that “two charged bodies will attract or repel one another with a force proportional to the product of their charges and inversely proportional to the square of the distance between them”,
Thus, its equation can be demonstrated as:
| \(\begin{array}{l} F ~= ~k \times \frac{Q_1 \times Q_2}{d^2} \end{array}\) |
Here,
- F = magnitude of force of attraction or repulsion
- k = Coulomb’s constant (Value of k = 9 × 109)
- q1 and q2 = magnitudes of two charges
- r = distance between two charges
The net force acting on a charge can be obtained by estimating the vector sum of all the forces acting on the charge. It is known as the Superposition Theorem.
Coulomb’s Law in Three-point charges
According to the example three-point charges QA, QB and QC with a position vector of r1, r2 and r3. Thus, the force the one charge experiences due to the other charges can be shown by:
\(\overrightarrow{F} = \overrightarrow{F_{AB}} + \overrightarrow{F_{BC}} + \overrightarrow{F_{CA}}\)
Thus, it can be written as:
| \(\begin{array}{l} \overrightarrow{F_1} ~=~ \mathop{\LARGE\mathrm \sum}_{j=1}^n F_{ij}\end{array}\), (herein, j ≠ i ) |
By applying this to the three-point charges:
⇒ \(\begin{array}{l} \overrightarrow{F_1} ~=~ \frac {1}{4 \pi \in} \left[ \frac {Q_A Q_B}{r^2_{AB}} \hat{r}_{AB} ~+~\frac {Q_A Q_C}{r^2_{AC}} \hat{r}_{AC} \right] \end{array}\)
Also Read:
| Important Topics Related to Coulomb’s Law | ||
|---|---|---|
| Unit of Electric Charge | Charge Density Formula | Electrical Insulators |
| Electric Field Physical Significance | Maxwell’s Equations | Dipole in Uniform External Field |
Coulomb's Law
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Coulomb’s law states that “force between two point charges varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges”.
As per the statement, the formula for force can be written as:
| \(\begin{array}{l} F ~= ~k \times \frac{Q_1 \times Q_2}{d^2} \end{array}\) |
Where,
- F is the magnitude of the force of attraction or repulsion depending on the charges
- k is the Coulomb’s constant
- q1 and q2 are the magnitudes of two charges
- r is the distance between two charges
- In SI units, the value of k is 9 × 109.
Coulomb’s Law
Also Read: Gauss's Law
- Coulomb determined the magnitude of the electric force by measuring the force between charged objects using a torsion balance (torsion balance is a device used to measure the gravitational acceleration on the Earth’s surface).
- Thus this force was experimentally proven among two static charged particles. Static means anything which does not move or change with time.
- In electrostatics, we generally use smaller units of charge like 1mC (10-3 C) or 1µC (10-6 C).
- Thus for ease, k is written as 1/4πεo. The modified Coulomb’s formula is written as:
F = \(\frac{1}{4\pi\epsilon_o}\) x \(\frac{q1*q2}{r*r}\)
Where,
- εo is the permittivity of free space (permittivity can be said as the ability of a material/ medium to permit or transmit electric field).
- In the SI unit, the value of εo is 8.854 × 10-12 C2 N-1 m-2.
The value of k thus depends on εo. If we understand how the force of charge between two point sources interacts as explained above, we will be able to easily understand how multiple charges exert force as explained in the upcoming topic.
Properties of Electric charge
The three qualities of an electric charge:
- Quantization: A body’s total charge denotes the integral multiple of a quantum of charge.
- Additive: This electric charge property represents the overall charge as the algebraic sum of all singular charges act on the system.
- Conservation: A system’s entire charge stays unchanged throughout time, as per conservation.
Force Between Multiple Charges Infograph
Also Read:- Electric Charge
Principle of Superposition
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The statement as experimentally verified is that the force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
Principle of Superposition
Coulomb’s law meets the principle of superposition, meaning that the force between two particles is not affected by the presence of other charges.
- This helps to determine the net force exerted on a charged particle by other charged particles.
- The force on a charged particle q0 due to the given charges q1, q2, and q3, can be expressed as the net force due to individual point charges.
- Thus, it can be said, \(\overrightarrow{F}_{0} = \overrightarrow{F}_{01} + \overrightarrow{F}_{02} + \overrightarrow{F}_{03} \)
Force is a vector quantity. Let us consider that there are three-point sources q1, q2, and q3 in a vacuum. To calculate the total force acting on q1 we need to add up the forces acting on q1 from q2 and q3. For force acting on q1 from q2:
For force acting on q1 from q3:
Thus the total force acting on q1 is:
As per the statement of the principle of superposition, the force exerted by q2 on q1 will be the same as per Coulomb’s law, that force is not affected by the presence of other point sources like q3, q4, …. qn. Thus, for calculating the total force ‘F1’ experienced by q1 we should add the vector sum of all the forces –
The parallelogram law of addition of vectors is used to obtain the vector sum of all the products. We can see here that it is the combination of both Coulomb's law as well as the principle of superposition to understand the forces acting from many charged sources on a single charged point.
Also Check:
Things to Remember
- When there are multiple charges, as per Coulomb’s law, it is considered that the force is always exerted over a straight line from the two centers of the point source or radially.
- The force experienced by a charged particle is directly dependent on the magnitude of the other charged source and inversely dependent on the distance separating the centres of the two charged point sources.
- When there are multiple charged point sources, the force on one point source adds up as the force from each point source acts individually on one charged point.
- The sign of the charge denotes its polarity and does not indicate its strength. The magnitude decides its strength.
- The positive or negative sign in the charges is used to understand the attraction or repulsion between charges.
- The charge of an electron is 1.6 × 10-19 C.
Also Read:- Electric Charges and Fields: Important Questions
Sample Questions
Ques. What is Coulomb's Law? (1 mark)
Ans. Coulomb’s law can be expressed as “two charged bodies attract or repel one another with a force proportional to the product of their charges and inversely proportional to the square of the distance between them”.
Ques. What are some key points of Coulomb’s Law? (4 marks)
Ans. Some key points of Coulomb’s Law include:
- Considering that the force between two charges that are in two different media is the same for different separations. Hence, \(\begin{array}{l}F=\frac{1}{K}\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} = constant\end{array} \)
- Kr2 = constant or K1r12 = K2r22
- In case the force between two charges separated, in vacuum, from each other by a distance ‘r0’ is equivalent to the force between the same charges now separated by a distance ‘r’ in a medium, then the Coulomb’s Law is: Kr2 = r02
- Two identical conductors that have q1 and q2 charges are connected. They have been further separated. Hence, each will have a charge equivalent to (q1 + q2)/2. However, if the charges are q1 and – q2, each will have a charge equal to (q1 – q2)/2.
Ques. Two similar-looking metallic spherical shells A and B with charges + 40 and -10 C have been placed quite some distance apart. A same and identical uncharged sphere C is first placed in contact with sphere A, then sphere B. Further, spheres A and B are brought into contact. Post that, they are separated. Determine the charge on the spheres A and B. (CBSE All India 2011 C) (2 marks)
Ans. When the two charges have been brought into contact, they get equally charged.
The charge’s magnitude = (q1 + q2)/2
Following the contact, the charge on sphere C with sphere A = (40+0)/2 = 20 C = Charge on sphere A
The Charge on C after coming in contact with B = (20-10)/2 = 5 C = Charge on sphere B
Thus, charge on spheres A and B = (20+5)/2
= 12.5 C on each sphere.
Ques. If 109 electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? (3 marks)
Ans. In one second 109 electrons move out of the body. Therefore the charge given out in one second is 1.6 × 10–19 × 109 C = 1.6 × 10–10 C.
The time required to accumulate a charge of 1 C can then be estimated to be
1 C ÷ (1.6 × 10–10 C/s)
= 6.25 × 109 s
= 6.25 × 109 ÷ (365 × 24 × 3600) years
= 198 years.
Thus to collect a charge of one coulomb, from a body from which 109 electrons move out every second, we will need approximately 200 years. One coulomb is, therefore, a very large unit for many practical purposes.
Ques. How much positive and negative charge is there in a cup of water? [3 marks]
Ans. Let us assume that the mass of one cup of water is 250 g. The molecular mass of water is 18g. Thus, one mole (= 6.02 × 1023 molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 1023. Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons. Hence the total positive and total negative charge has the same magnitude. It is equal to
(250/18) × 6.02 × 1023 × 10 × 1.6 × 10–19 C
= 1.34 × 107 C.
Ques. Two like and equal charges are at a distance of r = 5 cm and exert a force of F = 9 × 10-3 N on each other. [4 marks]
(a) Find the magnitude of each charge?
(b) What is the direction of the electrostatic force between them?
Ans. (a) Using Coulomb’s law,
F = k x \(\frac{q1*q2}{r*r}\)
As both the charges are equal, thus q1 = q2 = q.
We know that k is generally written as 9 × 109 Nm2/C2
As per SI unit system, the unit of r should be in meters thus r = 0.05 m … (because 1m=100cm)
Now solving for q:
F = k × (q2 / r2) … (both the charges are same and equal as given in the question)
9 × 10-3 = 9 × 109 × (q2 / (0.05)2)
q = 5 × 10-8 C
(b) As both the charges have the same polarity they have the same signs so the electric force between them is repulsive.
Ques. A point charge of q2 = 4µC is 3 cm apart from the charge q1 = -1μC [3 marks]
(a) Find the magnitude of the Coulomb force which one particle exerts on the other.
(b) Is the force attractive or repulsive?
Ans. Converting the given values into standard quantities
q1 = 1μC = -1 × 10-6
q2 = 4μC = 4 × 10-6
r = 3cm = 0.03 m
(a) Using Coulomb’s law,
F = k × (q2 / r2)
= 9 × 109 × [ (4 × 10-6) (-1 × 10-6) / (0.03)2]
= -40 N
(b) Since both the charges have opposite polarity, they will have attractive electric force between them.
Ques. Two charged particles apply an electric force of 5.2×10−3 N on each other. Distance between them is twice as much as before. What will be Coulomb's force? [3 marks]
Ans. Using Coulomb’s law,
F1 = k × (q1 × q2 / r12) … (1)
As per the question, r2 = 2r1 … (2)
Thus, after the distance gets twice as before, the new force will be,
F2 = k (q1 × q2 / r22)
= k (q1 × q2 / (2r1)2) … (from 2)
= 1/4 k (q1 × q2 / r12)
= 1/4 F1 … (from 1)
Thus, F2 = 1/4 (5.2×10−3)
F2 = 1.3 ×10−3 N
Ques. Two point charged particles are 4.41cm apart. They are moved and placed in a new position. The force between them is found to have been tripled. How far apart are they now? [3 marks]
Ans. Initial distance is r1=4.41cm. At the new location, the force is tripled F2 = 3F1. Applying Coulomb's law, we have
F2 = 3F1
k × (q1 × q2 / r22) = 3[k × (q1 × q2 / r12)]
1 / r22 = 3 / r12 … (cancelling similar quantities)
1 / r22 = 3 / (4.41 × 10-2 )2
r2 = 0.0254 m … (taking square root on both sides)
Thus, if those two charges are 2.54cm away, the electrostatic force between them gets tripled.
Ques. Four point charges are located on the corners of a square shown in the figure. If the net Coulomb force on q2 be zero, what is the ratio of Q/q? [5 marks]
Ans.
Since |q1| = |q3| = q and placed at an equal distance of the charge q2 so
F12 = F32.
We know that the resultant vector of two perpendicular and equal vectors F is given as √2F so, in this case, the magnitude of the net force acting on the charge q2 due to q1 and q3is
F = √2F12
along the diagonal (q2 − q4) of the square and directed outward as shown in the figure.
The total electric force on the charge q2 is the vector sum (superposition principle) of
F2 = F + F42
since said that it is F = 0 so the electrostatic force of q4 on q2 i.e. F42 must be equal in magnitude and opposite in direction with F. Therefore, by equating the magnitudes of the forces i.e. F = F42 we obtain
F = F42
√2 F12 = F42
√2 k × (q1 × q2 / r2) = k × (q4 × q2 / (√2r)2)
√2 |q| |Q| / 1 = |12Q| |Q| / 2
Q/q = 4√2
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