Electric Charges and Fields Class 12 | Chapter 1 | Physics | CBSE |
(I) Electrostatic Forces
The branch of physics which deals with the study of charge at rest (i.e., static charges), the force between the static charges, field and potential due to these charges is called Electrostatics.
Charge is something possessed by material object that make it possible for them to exert electrical force and to respond to electrical force. It is basic unit of an existing body and charge at rest is present only on insulating rubbed surface and that too for very short period of time.
- It is the energy with which electron are bounded with metal surface .
- It is measured in eV.
- Greater the work function, more will be energy required to remove an electron from metal surface.
- Platinum = Maximum work function
- Cesium = Minimum work function
Types of Charges
Electron : -vely charged ; Mass = 9.1 x 10-31 kg ; Charge = -1.6 x 10-19 Coulomb Proton : +vely charged ; Mass = 1.672 x 10-27 kg ; Charge = + 1.6 x 10-19 Coulomb Neutron : Electrically neutral ; Mass = 1.674 x 10-27 kg ; Charge = Zero
Properties of Charge
Additivity of charge is a property by virtue of which total charge of a system is obtained by simply adding algebraically all the charges present anywhere on the system.
Note : Proper sign have to be used while adding the charges in a system
Conservation of Charge
Conservation of charge is the property by virtue of which total charge of an isolated system always remain constant or conserved. It is not possible to create or destroy net charge carried by any isolated system.
Quantization of Electric Charge
The quantization of electric charge is the property by virtue of which, the charge on a body is integral multiple of a basic unit of charge of an electron / proton represented by e.
q = ± ne
Here, n is any integer The cause of quantization is that only integral number of electrons can be transferred from one body to another.
Question: What is the charge on a body from which one million electrons are moved ?
Solution : n = 106 , e = 1.6 x 10-19 C
As electrons are moved, charge acquired is positive
q = ne = 106 x 1.6 x 10-19 C = 1.6 x 10-13 C
The process of transferring of charge from one body to another body.
The transfer of charge between two bodies by their actual contact.
The transfer of charge between two bodies without their actual contact.
The transfer of charge between two bodies by rubbing them.
According to coulomb’s law, the force of attraction between any two bodies is directly proportional to the product of the charges and inversely proportional to square of distance between them.
F = k | q1 | | q2 | / r2
Here, k is electrostatic force constant. The value of electrostatic force constant depends on the nature of medium separating the charges, and on the system of units.
In SI Unit : k = 9 x 10-9 N m2 C-2
k = 1 / 4 π εo
εo is electrical permittivity of the free space
εo = 8.85 x 10-12 N-1 C2 m-2
Coulomb’s law is universal. The force which binds the electrons to the nucleus to form an atom is calculated from this law. It is Central Force as it acts along a line joining the two point two charges. It is a Conservative Force as the work is done in taking a charge from one point to another in the electric field of another charge is independent of path followed. It is not affected by presence or absence of any other charge.
These are kind of insulators which can conduct electricity when placed in external field.
Dielectric constant of a medium is the ratio of absolute electrical permittivity of the medium to the absolute electrical permittivity of free space.
Dielectric constant of a medium may be defined as the ratio of force of interaction between two point charges separated by a certain distance in air / vacuum to the force of attraction / repulsion between the same two point charges, held the same distance apart in the medium.
K = 1 for vacuum K = 1.003 ∼ 1 for air
Question : Two point charges having equal charges separated by distance 1 m distance experiences a force of 8 N. What will be the new force experienced by them, if they are held in water at the same distance ? ( K water = 80 )
Solution : F’ = F / K = 8 / 80 = 1 / 10 N
Coulomb’s Law in Vector Form
Let r 1 be position vector of q1
r 2 be position vector of q2
F12 be force on q1 due to q2
F21 be force on q2 due to q1
Thus, Coulomb’s Law is in accordance with Newton’s third law of motion.
Force Between Multiple Charges : Principle of Superposition
According to Superposition principle; total force on any charge due to a number of other charges at rest is the vector sum of all the forces on that charge due to other charges, taken one at a time. The forces due to individual charges are unaffected by the presence or absence of other charges.
Question : A charge of magnitude Q is divided into two parts q and (Q-q) such that the two parts exert maximum force on each other. Calculate the ratio Q / q
Solution: Force will be maximum only if dF / dq = 0
(1 / 4πεo) x d / dq [ q(Q-q) ] = 0
d / dq [ q(Q-q) ] = 0
q (-1) + (Q-q) = 0
Q – 2q = 0
Q = 2 q
Q / q = 2
Question : Two identical metal spheres A and B have equal and similar charges. They repel each other with a force of 103 N, when they are placed 10 cm apart in a medium of dielectric constant 7. Determine the charge on each sphere.
Solution : F = 1 / 4 π εo / K q2 r2
q = 28.6 x 10-6 C
Question : Force between two electrical charges kept at distance d apart in air is F. If the distance between two charges is halved and their individual charges are doubled then new forces between them will become.
Force due to Continuous Distribution of Charges
At the microscopic level, charge distribution is discontinuous as there are discrete charges separated by intervening space when there is no charge.
Linear Charge Distribution
Total force due to continuous line distribution of charge is as follows
Surface Charge Density
Total force due to continuous surface charge distribution is as follows
Volume Charge Density
By the superposition principle, total force due to entire volume charge density is obtained by summing over the forces due to different volume elements.
(II) Electrostatic Field
The Electric Field Intensity at any point is the strength of electric field at that point.
It is defined as the force experienced per unit positive charge placed at that point.
The charge Q which is producing the electric field is called a source charge and the charge qo which tests the effect of source charge is called a test charge.
Note : The test charge acts as detector of the electric field. It is an infinitely small charge so that it affects least the elctric field of source charge.
It is a vector quantity and direction of electric field intensity is along the direction of force. Its SI unit is N C-1
Electric Field Lines and Properties
An electric field line is a path, straight or curved in electric field, such that tangent to it at any point gives the direction of electric field intensity at that point.
Electric Field Lines due to a point positive charge
Electric Field Lines due to a point negative charge
Electric Field Lines due to an electric dipole
Electric Field Lines due to two equal positive charges
Electric Field Lines due to Positively charged plane conductor
Properties of Electric Lines of Force
- Electric Field lines are continuous curves. They start from a positively charged body and end at a negatively charged body.
- Electric field lines are continuous but do not form closed loops.
- Tangent to the electric field at any point at any point gives the direction of electric field intensity at that point.
- No two electric field lines can intersect each other because if they intersect each other then there will be two direction of electric field which is not possible. Moreover, they are considered with same charge and same charge never intersect each other.
- The electric field lines are always perpendicular to the surface of a conductor.
- The electric field lines contract longitudnally.
- The electric field lines exert a lateral pressure due to repulsion between the charges.
- Electric field lines travel from higher potential to lower potential.
- Electric field lines do not pass through conductor means inside a hollow conductor E = 0
- Closeness of electric field determines the strength of electric field.
Electric Field Intensity due to a Point Charge
Let Q be source charge
qo be a test charge
r be the separation between source charge and test charge
Electric Field intensity at a point depends only on source charge and separation between charge but not only on test charge
Electric Field Intensity on the Axis of Cicular Ring
Let a uniformly charged ring carrying charge q
R be radius of ring
We have to find electric field at point P which is at x distance from centre of ring
dq be small charge at ring which will produce dE electric field at point P.
dE at point P can be resolved as dEsinθ and dEcosθ ; dEsinθ cancel each other and dEcosθ at point P gives total effect at point P.
When x = 0 i.e., at centre of ring Enet = 0
Distance when electric Field is maximum
Electric Field intensity at centre due to circular arc
An electric dipole consists of a pair of equal and opposite point charges separated by some small distance.
The distance between the pair of charges or length of dipole is called as dipole length.
Note : Total force and total charge due to an electric dipole is zero but electric field associated with a dipole is not zero.
Dipole moment is a measure of the strength of electric dipole. It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and the distance between them.
P = q x 2a
By convention, the direction of P is from negative charge to positive charge.
SI unit of dipole moment is Coulomb metre.
Electric Field Intensity on Axial Line of an Electric Dipole
Let AB be an electric dipole consisting of two point charge q and -q separated by a small distance 2a. We have to calculate electric field intensity E at a point P on the axial line of the dipole and at a distance x from the centre of dipole
1 ) If x >>> a; a2 is neglected ( Very short dipole)
Enet = 2KPx / (x2)2 = 2 KP / x3
This is the required expression for electric field intensity on axial line due to short dipole
2 ) If whole system is placed in a medium of dielectric constant (K)
E axial (medium) = Eaxial / K
E axial (medium) = 2 KP / x3 / K
Electric Field Intensity on Equatorial Line of an Electric Dipole
Consider an electric dipole consisting of two point charges -q and +q separated by a small distance AB = 2a with a centre at O.
We have to find electric field intensity E at point P on the equatorial line of the dipole OP = r
E can be resolved as Esinθ and Ecosθ
EA sinθ and EB sinθ cancel each other while EA cosθ and EB cosθ gives the net effect.
ENet = EA cosθ + EB cosθ
If x>>> a ; a2 is neglected
When whole system of charges is placed in a medium whose dielectric constant is K
Eeq. (medium) = Eeq. / K
Relation between Eaxial and Eeq. for a short dipole :
Electric Field Intensity due to an Electric Dipole at some General Point
Consider an electric dipole consisting of two point charges -q and +q separated by a small distance AB = 2a with a centre at O.
Let E1 be field intensity on the axial line and E2 be field intensity on equatorial point
Torque Experienced by an Electric Dipole into a Uniform Electric Field
Let E be uniform electric field
P be Electric dipole moment
±q be pair of charges
2a be dipole length
Torque = Force x Perpendicular displacement
Torque = ± qE x 2a sin θ
Torque = (q x 2a) E sin θ
Torque ; τ = P E sin θ
Special Cases For Torque on an Electric Dipole
1) When Electric dipole is parallel to the electric field .i.e., P II E
θ = 0°
τ = PE sin 0°
τ = 0 (minimum)
2) When Electric dipole is perpendicular to the electric field
θ = 90°
τ = PE sin 90°
τ = PE ( maximum )
3) When Electric dipole is anti-parallel to the electric field .i.e., P II E
θ = 180°
τ = PE sin 180°
τ = 0 (minimum)
Electric Potential Energy associated with an Electric Dipole placed in Uniform Electric Field
Let θ1 be initial angle
θ2 be final angle
dθ be angular displacement
1) If θ1 = 0° and θ2 = θ
W = -PE [ cos θ2 – cos 0]
W = -PE [ cos θ2 – 1]
2) If θ1 = 90° and θ2 = 0°
W = -PE [ cos 0° – cos 90°]
W = -PE (1)
W = -PE (Minimum energy) (Stable condition )
3) If θ1 = 90° and θ2 = 180°
W = -PE [ cos 180° – cos 90°]
W = -PE (-1)
W = PE (Maximum energy) (Unstable condition )
Question : Calculate the amount of work done required to rotate an electric dipole from stable to unstable condition into a uniform electric Field.
Solution : W = -PE [ cos θ2 – cos θ1 ]
In stable condition ; θ1 = 0°
In Unstable condition ; θ2 = 180°
W = -PE [ cos 180° – cos 0°]
W = -PE [-1-1]
W = 2 PE
Question : An electric dipole of length 2 cm when placed with its axis making an angle of 60° with a uniform electric field experiences a torque of 8√3 N m. Calculate the potential energy of the dipole if it has a charge of ± 4 nC.
Solution : Length (2a) = 2 x 10-2 m
Angle, θ = 60°
Torque = 8√3 Nm
Charge = 4 x 10-9 C
We know; τ = Q (2a) E sinθ
E = 8√3 / 4 x 10-9 x 2 x 10-2 x sin 60°
E = 2 x 10 11 N C-1
Now; U = -PE cosθ = – 4 x 10-9 x 2 x 10-2 x 0.5 x 2 x 10 11 = – 8 J
Question : Two point charges of ±3.2 x 10-19 C separated by 2.4 x 10-10 m . The dipole is placed in a uniform electric field of 4 x 105 N C-1. Calculate work done in rotating dipole by 180°.
Solution : q = ± 3.2 x 10-19 C
E = 4 x 105 N
2a = 2.4 x 10-10 m
We know; work done in rotation the dipole = 2 PE
U = 2 x 3.2 x 10-19 x 2.4 x 10-10 x 4 x 105
U = 61.44 x 10-24 J
Force associated with an Electric Dipole in Non Uniform Electric Field
Let ±q be pair of charges 2a be dipole length
(III) Electric Flux
By convention , the vector associated with every area element of a closed surface is taken to be in the direction of the outward drawn normal.
Electric flux over an area in an electric field is measure of the number of electric field lines crossing this area.
Electric Flux is a scalar quantity. Electric flux over a closed surface can be positive, negative or zero.
Its SI unit is Nm2C-2
Dimensional formula of ΦE is [ML3T-3A-1]
The surface integral of electric field E produced by any source over any closed surface S enclosing a volume V in vacuum i.e., total electric flux over a closed surface S in vacuum is 1 / εo times the total charge (Q) contained inside S.
Note : There is no contribution to electric flux from the charge outside S. Further, the location of Q inside S does not affect the value of surface integral.
The charge inside S may be point charge or even continuous charge distribution. The surface chosen to calculate the surface integral is called Gaussian surface.
Proof of Gauss’s theorem (For spherically symmetrical surfaces only)
Suppose an isolated positive point charge is situated at centre O of a sphere of radius r.
Note: Electric Flux does not depend on shape of surface but it depends on the charge placed inside the closed body. Moreover, the surface need not to be a real physical surface; it can also be a hypothetical one.
Some Important cases related to Gauss’s law
| Φin = -πR2E
Φout = πR2E
Φtotal = 0
|Φin = Φcircular = -πR2E
Φout = Φcurved = πR2E
Φtotal = 0
|Φin = -1/2 πR2E
Φout = 1/2 πR2E
Φtotal = 0
|Note : Here Electric Field is radial
Φ = q / 2 εo
|Φ = q / 2 εo|
Electric Flux for a Cube at Different Locations
If a charge is at centre of cube
ΦTotal = Q / εo
ΦFace = Q / 6 εo
ΦCorner = Q / 8 εo
ΦEdge = Q / 12 εo
Question: A surface S1 of radius r1 encloses in a charge Q. If there is another concentric sphere S2 (r2 > r1) enclosing charge 2Q. Find the ratio of electric flux through S1 and S2. How will the electric flux through sphere S1 change if a medium of dielectric constant K is introduced in the space inside S2 in place of air?
Solution: According to Gauss’s law;
Flux through S1 = Q / εo
Flux through S2 = Q + 2Q / εo
On dividing equation (i) by equation (ii) , we get
Φ1 / Φ2 = 1 / 3
There is no change in electric flux through S1 with dielectric medium inside the spare S2.
Deduction of Coulomb’s Law from Gauss’s Law
Consider an isolated positive point charge q at O. Imagine a sphere of radius r with centre O. The magnitude of electric field intensity E at every point on the surface of the sphere is the same and it is directed radially outwards. Further, the direction of vector ds representing a small area element ds on the surface of the sphere is along E only, i.e., θ = 0°.
This is the electric field intensity at point P distant r from an isolated point charge q at the centre of the sphere. If another point charge qo were placed at P, then force on qo would be
This is Coulomb’s Law
Important Application of Gauss’s Theorem
Electric Field Intensity due to an Infinitely Long Straight Wire
Consider an infinitely long thin wire with uniform linear charge density λ. In order to find electric field at distance r from it; there is a gaussian surface.
This cylinder has four surfaces. 1,4 are plane surfaces and 2,3 are curved cylindrical surfaces.
Let Φ1, Φ2, Φ3 , Φ4 are flux associated at 1,2,3,4 surfaces.
L be length of cylinder upto gaussian surface.
Electric Field Intensity due to uniformly charged Spherical Shell
Consider a thin spherical shell of radius R with centre O. Let a charge +q be distributed uniformly over the surface of the shell. The surface of this sphere is Gaussian surface at every point of which electric field intensity E is the same, directed radially outwards
Electric Field Intensity at a point outside the shell ( r > R)
Electric Field Intensity at a point on the surface of shell ( r = R)
Electric Field Intensity at a point inside the shell
We know, in case of spherical shell; no charge exists inside the gaussian surface i.e., E = 0
Variation of Electric Field Intensity (E) with distance (r) from the centre O of a charged spherical shell
Electric Field Intensity due to a Non – Conducting Charges Solid Sphere
Suppose, a non conducting solid sphere of radius R and centre O has uniform volume density of charge ρ.
Electric Field Intensity at a point outside the solid sphere ( r > R )
Draw a gaussian surface in the form of a sphere of radius r with O as centre. From Gauss’s law ;
Electric Field Intensity at a point on the surface of solid sphere ( r = R)
Electric Field Intensity at a point inside the solid sphere
Draw a gaussian surface of radius r with O as same charge enclosed by gaussian surface
Variation of Electric Field Intensity (E) with distance (r) from the centre O of a non conducting charged solid sphere
Electric Field Intensity due to Thin Infinite Plane Sheet of Charges
Consider a thin infinite plane sheet of charge. Let σ be the surface density of charge on the sheet.
From symmetry, we find that E on either side of the sheet must be perpendicular to the plane of the sheet, having the same magnitude at all points equidistant from the sheet.
Let us imagine a cylinder of cross sectional area ds piercing through the sheet.
If the sheet carries positive charge, σ > 0. The electric field is uniform , normal and outwards from the sheet. If the sheet carries σ < 0 ; the field is uniform and along the inward normal to the plane sheet.
Electric Field Intensity due to Two Thin Infinite Plane Parallel Sheets of Charges
Let A and B be two infinite plane charged sheets held parallel to each other.
σ1 = Uniform surface density of charge on A
σ2 = Uniform charge density of charge on B
Question : A hollow cylindrical box of length 1 m and area of cross section 25 cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by E = 50 xî, where E is in NC-1 ans x is in metre. Find net electric flux through the cylinder.
Solution : Given, E = 50 xî
As the flux is only along the X axis, so flux will pass only through the cross sectional area of the cylinder
Magnitude of electric field at cross section A, EA = 50 x 1 = 50 N C-1
Magnitude of electric field at cross section A, EB = 50 x 2 = 100 N C-1
The corresponding electric fluxes are
ΦA = EA. ΔS = 50 x 25 x 10-4 x cos 180° = -0.125 Nm2C-1
ΦB = EB. ΔS = 100 x 25 x 10-4 x cos 0° = 0.25 Nm2C-1
So net flux through the cylinder, Φ = ΦA + ΦB =-0.125+0.25
Φ = 0.125 Nm2C-1
Question : Given a uniform electric field E = 5 x 103 î NC-1, find the flux of this field through a square of 10 cm on a side whose plane is parallel to YZ plane. What would be flux through the same square if the plane makes an angle of 30° with X axis.
Solution: Given; electric field intensity = 5 x 103 î NC-1
Side of square, S = 10 cm = 0.1 m
Area of square = (0.1)2 = 0.01 m2
The plane of this square is parallel to YZ plane. Hence, the angle between the unit vector normal to the plane and electric field is zero i.e., θ = 0°
Flux through the plane, Φ = E.A cosθ = 5 x 103 x 0.01 cos 0° = 50 Nm2C-1
If the plane makes an angle of 30° with X axis then θ = 60°
Flux through the plane, Φ = E.A cosθ = 5 x 103 x 0.01 cos 60° = 25 Nm2C-1
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