Many of us have experienced hearing a crackle when we take off our synthetic shirts or nylon sweaters. Sometimes, we also feel the sensation of an electric shock while opening the door of our car. The reason for these experiences is the discharge of electric charges through our body, which accumulate due to rubbing of insulating surfaces. Another common example of electric discharge is the lightning that we see in the sky during thunderstorms.
In this document, we will study in detail about the concepts of electrostatics.
Electrostatics is a branch of physics that deals with the study of stationary electric charges and their interactions. It focuses on understanding the behavior of electric charges at rest and the forces that arise between them.
Electrostatic deals with Static Charge
study of electric and magnetic interactions. and Electrostatics is the branch of electromagnetics dealing with the effects of electric charges at rest
The property associated with matter due to which it produces and experiences electric and magnetic fields is Electric Charge. The SI unit of charge is coulomb (C) and e has the value
e = 1.6 * 10-19C
Properties of Electric Charges
Regarding charge the following points are worth noting:
Q.1. How many electrons are there in one coulomb of negative charge?
Answer: The negative charge is due to the presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is e = 1.6×10-19C , the number of electrons is equal to the charge q divided by the charge e on each electron. Therefore, the number n of electrons is
Mainly there are the following three methods of charging a body :
- Conduction
- Induction
- Friction
Charging a body by friction
Q.2. If we comb our hair on a dry day and bring the comb near small pieces of paper, the comb attracts the pieces, why?
Answer: This is an example of frictional electricity and induction. When we comb our hair, it gets positively charged by rubbing. When the comb is brought near the pieces of paper some of the electrons accumulate at the edge of the paper piece which is closer to the comb. At the farther end of the piece there is deficiency of electrons and hence, positive charge appears there. Such a redistribution of charge in a material, due to presence of a nearby charged body is called induction. The comb exerts larger attraction on the negative charges of the paper piece as compared to the repulsion on the positive charge. This is because the negative charges are closer to the comb. Hence, there is a net attraction between the comb and the paper piece.
Coulomb's Law states that the electrostatic force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and varies inversely as the square of the distance between the two bodies.
Coulomb's Law
Mathematically, it can be expressed as:
F = k * (|q1 * q2|) / r2
Where:
Regarding Coulomb's law, the following key points are worth noting:
Superposition Principle
Q.3. What is the smallest electric force between two charges placed at a distance of 1.0 m?
When two charged particles are at a distance from each other, they cannot directly interact. However, a charged particle creates an electric field in the space around it. This electric field exerts a force on any other charge placed within it, except for the source charge itself.
Electric Field
The strength of an electric field is measured by the force experienced by a unit positive charge placed at that point. The direction of the field is given by the direction of motion of a unit positive charge if it were free to move.
Electric Field Strength Formula
For positive source charge, the electric field is radially outward, whereas, for negative source charge, the electric field is radially inward.
Unit of Electric field
E = [Newton/Coulomb] or [Joule/(Coulomb) (meter)]
An Electric Field Leads to a Force:Suppose there is an electric field strength E at some point in an electric field, then the electrostatic force acting on a charge +q is qE in the direction of E, while on the charge – q it is qE in the opposite direction of E.
Q.4. An electric field of 105 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of + 2 μC and − 5 μC at this spot?
Solution:
The strength and direction of the electric field are represented by electric lines of forces or electric field lines.
Electric Field Lines
Characteristics of Electric Field Lines:
Properties of Electric Field Lines:
Electric Field Lines
Q.5. Two point charges +8q and -2q are located at x = 0 and x = L respectively. The point on x axis at which net electric field is zero due to these charges is-
Solution: By principle of superposition, the Electric field at a point will be the sum of electric field due to the two charges +8q and -2q∴ 2x - 2L = x
∴ x = +2L
A conducting ring of radius R has a total charge q uniformly distributed over its circumference. We are interested in finding the electric field at point P that lies on the axis of the ring at a distance x from its centre.
Formula of Ex (Electric Field of a Ring of Charge)
Positive charge q is distributed uniformly along a line with length 2a, lying along the y-axis between
y = –a and y = +a . We are here interested in finding the electric field at point P on the x-axis.
Formula of Ex (Electric Field of a Line Charge)
that λ, the charge per unit length remains constant. In this case, Equation can be written as:
Consider an electric field generated by a positive point charge. The direction of the electric field is such that it is radially outwards.
Two electric charges, interacting with each other due to their electric field, have the electric potential energy defined as:
U = (k · q1 · q2) / r
where:
The Coulomb constant also known as the electric force constant, or the electrostatic constant, has a fixed value, equal to:
The unit of measurement of electric potential energy is joule [J].
Q.6. Two charges of magnitude 2 nC and 3 nC are placed at 2 cm from each other. Calculate the electric potential energy between these two charges.
Solution:
Given
q1 = 2 nC
q2 = 3 nC
r = 2 cm = 0.02 m
The electric potential energy is given by
U = kq1q2/r
Or, U = (9 x 109 Nm2C-2 x 2 x 10-9 C x 2 x 10-9 C)/0.02 m
Or, U = 2.7 x 10-6J
W = U = 1/(4πε0) (q1q2/r12) = q1V1
Q.7. A particle with the charge of -5 nanocoulombs is at a distance of 10 centimeters away from another charge of 10 nanocoulombs. Calculate the potential energy of the systems form by these two electric charges.
Solution:
Step 1. Convert the electrical charges from [nC] to [C] by multiplying the [nC] value with 10-9:q1 = -5 · 10-9 C
q2 = 10 · 10-9 C
Step 2. Convert the distance from [cm] to [m] by dividing the [cm] value to 100:
r = 10 /100 = 0.1 m
Step 3. Calculate the electric potential energy of the system using equation:
U = (k · q1 · q2) / rU = (8.9875517923 · 109 · (-5) · 10-9 · 10 · 10-9) / 0.1 = – 4.494 · 10-6 J
The resultant electric potential energy is negative, which means that there is an attraction force between the two particles. This is valid since one particle has positive charge and the second particle has negative charge.
W = U = (1/4πε0) (q1q2/r12 + q1q3/r13 + q2q3/r23)
The total energy of this system can be obtained by adding the potential energies of each possible pair.
W = U = (1/4πε0) (q1q2/r12 + q2q3/r23 + q3q4/r34+ q4q1/r41+ q1q3/r13 + q2q4/r24)
Potential energy of an electric dipole, in an electrostatic field, is defined as the work done in rotating the dipole from zero energy position to the desired position in the electric field.
Electric Potential
Electric potential, at any point, is defined as the negative line integral of the electric field from infinity to that point along any path.
The potential difference, between any two points, in an electric field, is defined as the work done in taking a unit positive charge from one point to the other against the electric field.
WAB = q [VA-VB]
So, V = [VA-VB] = W/q
Units & Dimension
Q.8. A +2 μC test charge is initially at rest a distance of 15 cm from a +7 μC source charge fixed at the origin. The Coulomb force pushes the test charge away from the source charge, reaching 20 cm. What is the work done by the electric field?
Solution:
Given
q1 = +2 μC
q2= +7 μC
rA = 15 cm = 0.15 m
rB = 20 cm = 0.20 m
The work done is given by
W = -ΔV
Or, W = -(kq1q2)(1/rB – 1/rA)
Or, W = -(9 x 109 Nm2C-2 x 7 x 10-6 C x 2 x 10-6 C)(1/0.20 m- 1/0.15 m)
Or, W = 0.21 J
V = (1/4π ε0) (q/r)
V = (1/4π ε0) [q1/r1 + q2/r2 + q3/r3] = V1+V2+ V2+….
Common potential, V = (1/4π ε0) [(Q1+Q2)/(r1+r2)]
q1 = r1(Q1+Q2)/(r1+r2) = r1Q/ r1+r2
q2 = r2Q/ r1+r2
q1/q2 = r1/r2 or σ1/ σ2 = r1/r2
V (r,θ) = qa cosθ/(4πε0r2) = p cosθ/(4πε0r2)
If n drops coalesce to form one drop, then,
K. E = 1/2 mv2 = eV
An electric dipole is defined as a couple of opposite charges q and –q separated by a distance d.
Dipole moment () of an electric dipole is defined as the product of the magnitude of one of the charges and the vector distance from negative to positive charge.
Unit of Dipole Moment:
Consider an electric dipole lying along positive y-direction with its centre at origin.
The electric potential due to this dipole at point A x y z ( , , ) as shown is simply the sum of the potentials due to the two charges. Thus,
Electric Potential and Field Due to an Electric DipoleSpecial Cases:
1. On the axis of the dipole:
2. On the perpendicular bisector of dipole
As there are too many formulae in electric dipole, we have summarised them as under :
1. | p | = (2a) q
Direction of p is from −q to + q.
2. If a dipole is placed along y-axis with its centre at origin, then
3. On the axis of dipole x = 0, z = 0
4. On the perpendicular bisector of dipole Along x-axis, y = 0, z = 0
5. Dipole in uniform electric field
Solution:
Q.10. Along the axis of a dipole, direction of electric field is always in the direction of electric dipole moment p. Is this statement true or false?
False. In the above figure, we can see that direction of electric field is in the opposite direction of p between the two charges.
Q.11. At a far away distance r along the axis from an electric dipole electric field is E. Find the electric field at distance 2r along the perpendicular bisector.
This law gives a relation between the net electric flux through a closed surface and the charge enclosed by the surface. According to this law, “the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε0 .” In symbols, it can be written as
where, qin represents the net charge inside the closed surface and Erepresents the electric field at any point on the surface.
but this form of Gauss’s law is applicable only under the following two conditions:
(i) The electric field at every point on the surface is either perpendicular or tangential.
(ii) Magnitude of electric field at every point where it is perpendicular to the surface has a constant value (say E).
As Gauss’s law does not provide expression for electric field but provides only for its flux through a closed surface. To calculate E we choose an imaginary closed surface (called Gaussian surface) in which Simplified Form of Gauss's Theorem can be applied easily. Let us discuss few simple cases.
1. Electric field due to a point charge
Electric field due to a point charge
2. Electric field due to a linear charge distribution
Let us consider an infinitely long wire with linear charge density λ and length L. To calculate electric field, we assume a cylindrical Gaussian surface. As the electric field E is radial in direction, the flux through the end of the cylindrical surface will be zero.
This is because the electric field and area vector are perpendicular to each other. As the electric field is perpendicular to every point of the curved surface, we can say that its magnitude will be constant.
The surface area of the curved cylindrical surface is 2πrl. The electric flux through the curve is E × 2πrl
According to Gauss’s Law
You need to remember that the direction of the electric field is radially outward if linear charge density is positive. On the other hand, it will be radially inward if the linear charge density is negative.
3. Electric Field due to Infinite Plate Sheet
Let us consider an infinite plane sheet, with surface charge density σ and cross-sectional area A. The position of the infinite plane sheet is as below:The direction of the electric field due to an infinite charge sheet is perpendicular to the plane of the sheet. Let us consider a cylindrical Gaussian surface, whose axis is normal to the plane of the sheet. We can evaluate the electric field E from Gauss’s Law as according to the law:
From a continuous charge distribution charge q will be the charge density (σ) times the area (A). Talking about net electric flux, we will consider electric flux only from the two ends of the assumed Gaussian surface. We can attribute it to the fact that the curved surface area and an electric field are normal to each other, thereby producing zero electric flux. So the net electric flux is
Φ = EA – (– EA)
Φ = 2EA
Then, we can write
The term A cancels out which means electric field due to an infinite plane sheet is independent of cross-sectional area A and equals to:
4. Electric Field due to Thin Spherical Shell
Let us consider a thin spherical shell of surface charge density σ and radius “R”. By observation, we can see that the shell has spherical symmetry. Therefore, we can evaluate the electric field due to the spherical shell in two different positions:
Electric Field Outside the Spherical Shell
To find electric field outside the spherical shell, we take a point P outside the shell at a distance r from the centre of the spherical shell. By symmetry, we take Gaussian spherical surface with radius r and centre O. The Gaussian surface will pass through P, and experience a constant electric field E all around as all points are equally distanced “r’’ from the centre of the sphere. Then, according to Gauss’s Law:
The enclosed charge inside the Gaussian surface q will be σ × 4 πR2. The total electric flux through the Gaussian surface will be
Φ = E × 4 πr2
Then by Gauss’s Law, we can write
Putting the value of surface charge density σ as q/4 πR2, we can rewrite the electric field as
In vector form, the electric field is
where r is the radius vector, depicting the direction of electric field. What we must note here is that if the surface charge density σ is negative, the direction of the electric field will be radially inward.
Electric Field Inside the Spherical Shell
To evaluate electric field inside the spherical shell, let’s take a point P inside the spherical shell. By symmetry, we again take a spherical Gaussian surface passing through P, centered at O and with radius r. Now according to Gauss’s Law
The net electric flux will be E × 4 π r2.
Q.12. A uniform electric field of magnitude E = 100 N/C exists in the space in the X-direction. Using the Gauss theorem, calculate the flux of this field through a plane, square area of edge 10 cm placed in the Y-Z plane. Take the normal along the positive X-axis to be positive.
Solution:
The flux Φ = ∫ E.cosθ ds
As the normal to the area points along the electric field, θ = 0
Also, E is uniform so, Φ = E.ΔS = (100 N/C) (0.10m)2 = 1 N-m2
Q.13. A large plane charge sheet having surface charge density σ = 2.0 × 10-6 C-m-2 lies in the X-Y plane. Find the flux of the electric field through a circular area of radius 1 cm lying completely in the region where x, y and z are all positive and with its normal, making an angle of 600 with the Z-axis.
Solution: The electric field near the plane charge sheet is E = σ/2ε0 in the direction away from the sheet. At the given area, the field is along the Z-axis.
The area = πr2 = 3.14 × 1 cm2 = 3.14 × 10-4 m2.
The angle between the normal to the area and the field is 600.
Hence, according to Gauss theorem, the flux
Q.14. A particle of mass 5 × 10-6g is kept over a large horizontal sheet of charge of density 4.0 × 10-6 C/m2 (figure). What charge should be given to this particle so that if released, it does not fall down? How many electrons are to be removed to give this charge? How much mass is decreased due to the removal of these electrons?
Solution:The electric field in front of the sheet is
E = σ/2ε0 = (4.0 × 10-6)/(2 × 8.85 × 10-12) = 2.26 × 105 N/C
If a charge q is given to the particle, the electric force qE acts in the upward direction. It will balance the weight of the particle, if
q × 2.26 × 105 N/C = 5 × 10-9 kg × 9.8 m/s2
or, q = [4.9 × 10-8]/[2.26 × 105]C = 2.21 × 10-13 C
The charge on one electron is 1.6 × 10-19C. The number of electrons to be removed
= [2.21 × 10-13]/[1.6 × 10-19] = 1.4 × 106
Mass decreased due to the removal of these electrons = 1.4 × 106 × 9.1 × 10-31 kg = 1.3 × 10-24 kg.
At all points inside the charged spherical conductor or hollow spherical shell, electric field E = 0, as there is no charge inside such a sphere.
Electric Field:
Potential:
Electric Field:
Potential:
The capacitance of a conductor is defined as the ratio between the charge of the conductor to its potential.
C = Q/V
Units & Dimension
C = 4πε0r
Capacitor: A capacitor or a condenser is an arrangement that provides a larger capacity in a smaller space.
Cair = ε0A/d
Cmed = Kε0A/d
Here, A is the common area of the two plates and d is the distance between the plates.
C = ε0A/[d-t+(t/K)]
Here d is the separation between the plates, t is the thickness of the dielectric slab, A is the area and K is the dielectric constant of the material of the slab.
If the space is completely filled with dielectric medium (t=d), then, C = ε0KA/ d
Increase in capacity, ΔC = 4π ε0b
It signifies, by connecting the inner sphere to earth and charging the outer one we get an additional capacity equal to the capacity of the outer sphere.
W = ½ QV = ½ Q2/C = ½ CV2
U = ½ ε0E2 = ½ (σ2/ ε0)
This signifies the energy density of a capacitor is independent of the area of plates and the distance between them so, the value of E does not change.
Capacitors in parallel:
Capacitors in series:
Energy stored in a group of capacitors:
(a) F = ½ ε0E2A
(b) F = σ2A/2ε0
(c) F=Q2/2ε0A
F = (Q2/2C2) (dC/dx) = ½ V2 (dC/dx)
V = [C1V1+ C2V2] / [C1+C2] = [Q1+Q2]/ [C1+C2]
ΔQ = [C1C2/C1+C2] [V1-V2]
ΔU = ½ [C1C2/C1+C2] [V1-V2] 2
At a point on its axis, E = (1/4πε0)(2πσ)[1 - x/(a2+R2)1/2] Here, σ is the surface charge.
Here, Q is the total charge
E = σ/2ε0
E = σ/2ε0
This signifies, the electric field near a charged sheet is independent of the distance of the point from the sheet and depends only upon its charge density and is directed normally to the sheet.
E= σ/ε0
Pelec = (1/2 ε0) σ2
357 docs|148 tests
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1. What is Electrostatics? |
2. What is an Electric Charge? |
3. What is Coulomb's Law? |
4. What is Electric Potential? |
5. What is Gauss's Law? |
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