Electric Charge & Field Notes

Electric Charge (q) & Field

Electric Charge (q)

• Intrinsic property of matter that experiences or exerts electric force.

• Measured in Coulombs (C).

Point Charge

• A charged body whose size is significantly smaller than its distance from surrounding objects.

Types of Charges (q)

1. Positive Charge: Protons

2. Negative Charge: Electrons

SI Unit of Charge

Basic Properties of Electric Charges

1. Additive Nature of Electric Charge

2. Conservation of Charge

3. Quantization of Charge

1. Additive Nature of Electric Charge

• Electric charge is additive; the total charge on a body is the algebraic sum of all charges present.

• If a system contains charges $q<em>1$ and $q</em>2$, the total charge is: $q<em>1 + q</em>2$.

Example

• Sphere A has a charge of -50 C, and sphere B has a charge of +20 C.

• The spheres are identical in size and made of conducting materials.

• If the spheres are touched together, the resulting charge on each sphere will be the average of the two charges: $(-50 + 20)/2 = -15 C$.

2. Conservation of Charge

• The total amount of charge in an isolated system remains constant.

• Charges are neither created nor destroyed but transferred from one system to another.

3. Quantization of Charge

Question: How many electrons are there in 1 C of charge?

• Total charge, $Q = ne$

• $n = Q/e$ where $e = 1.6 \times 10^{-19} C$

• $n = \frac{1 C}{1.6 \times 10^{-19} C} = 6.25 \times 10^{18}$ electrons

• $6.25 \times 10^{18}$ electrons constitute 1 C of charge.

Question

• Two identical neutral gold coins. How many electrons should be transferred to one coin to make them attract each other, if the charge supplied is 3.2 C?

• $Q = ne \Rightarrow n = \frac{Q}{e}$

• $n = \frac{3.2 C}{1.6 \times 10^{-19} C} = 2.0 \times 10^{19}$ electrons

• Therefore, $2.0 \times 10^{19}$ electrons have to transfer to give a charge of +3.2 C.

Coulomb's Law

• The magnitude of the electrostatic force of repulsion or attraction between two stationary point charges $q<em>1$ and $q</em>2$ is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance $r$ between them.

• $F \propto q<em>1 q</em>2$ ---(i)

• $F \propto \frac{1}{r^2}$ ---(ii)

Combining the above

• $F \propto \frac{q<em>1 q</em>2}{r^2}$

• $F = K \frac{q<em>1 q</em>2}{r^2}$, where K is Coulomb's constant.

• $K = \frac{1}{4 \pi \epsilon<em>0} = 9 \times 10^9 Nm^2C^{-2}$, and $\epsilon</em>0$ is the permittivity of free space (air or vacuum) with a value of $8.85 \times 10^{-12} C^2/Nm^2$.

• Note: Coulomb's law applies only to point charges and is also called the inverse square law.

Dielectrics

• Materials that do not allow current to flow.

• They are insulating materials such as glass, paper, mica, etc.

• Often called insulators because they are the opposite of conductors.

Relative Permittivity ($\epsilon_r$) or Dielectric Constant (K)

• The ratio of the permittivity of a medium ($\epsilon$) to the permittivity of free space ($\epsilon_0$).

• It is dimensionless.

• For example, water has a dielectric constant, $K \approx 80$.

Formula

• $K = \frac{\epsilon}{\epsilon_0}$

• $\epsilon = K \epsilon_0$

Coulomb's Law in Air

• $F = \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q<em>2}{r^2}$, where $\epsilon</em>0$ is the permittivity of free space or vacuum.

Coulomb's Law in a Medium

• $F = \frac{1}{4 \pi \epsilon} \frac{q<em>1 q</em>2}{r^2}$, where $\epsilon$ is the permittivity of the medium.

• $F = \frac{1}{4 \pi K \epsilon<em>0} \frac{q</em>1 q<em>2}{r^2}$ since $\epsilon = K \epsilon</em>0$

In terms of dielectric constant

• $F = \frac{1}{K} \frac{q<em>1 q</em>2}{4 \pi \epsilon_0 r^2}$

Question

• In comparison to the electrostatic force between two electrons, the electrostatic force between two protons is the same because the magnitude of charge of electron and proton is the same, $e$.

Graph

• The graph between F and $1/r^2$ is a straight line of slope $\frac{1}{4 \pi \epsilon<em>0} q</em>1 q_2$ passing through the origin.

Question

• Identify which of the graphs represents the force between two electrons vs. distance between them. (Graph A)

Relate Force in air and in medium of dielectric constant, K.

• If there are some dielectrics between the charges, then the force between them decreases by a constant factor K.

• The dielectric constant of water is about 80, and for a conductor, it is infinity ($\infty$).

Define one coulomb of charge.

Question

• The electrostatic force between the two charges $q<em>1 = 26 \mu C$ and $q</em>2 = -47 \mu C$ is 5.7 N. Calculate the separation between $q<em>1$ and $q</em>2$.

• If the medium between the charges is not free space, then the electrostatic force will decrease.

Question

• Two spheres each of q C are separated by a distance r meter. If the charges are doubled and the separation is halved, calculate the force of repulsion.

• $F = K \frac{q q}{r^2}$ ---(1)

• Now if new charges are 2q C each and separation is then $r/2$ then

• $F' = K \frac{2q 2q}{(r/2)^2} = K \frac{4q^2}{r^2/4} = 16 K \frac{q^2}{r^2} = 16 F$

• Hence, the force of repulsion becomes 16 times the initial force.

Forces due to multiple charges (Principle of Superposition)

• According to the principle of superposition, the total force acting on a charged particle (say $q_1$) due to ‘n’ number of particles surrounding it is equal to the vector sum of forces exerted by the individual charged particles.

Discuss

• The figure shows two protons (p) and one electron (e) on an axis.

  • What is the direction of electrostatic force

    • (a) on the central proton due to the electron? Towards the electron due to attraction (i.e., towards left).

    • (b) On the central proton due to another proton? Towards the electron due to repulsion (i.e., towards the left).

    • (c) Net force on the central force? Towards the electron (i.e., towards the left).

Question Two point charges +2 μC and +6 μC repel each other with a force of 12 N. if each is given an additional charge of -4 μC, what will be the new force between them?

• $q_1$ = +2μC=+2x10⁻⁶ C

• $q_2$ = +6μC=+6x10⁻⁶ C

• $F = K \frac{q<em>1 q</em>2}{r^2}$

• $12 = K \frac{(2x10^{-6})(6x10^{-6})}{r^2}$

• $r^2 = K \frac{(2x10^{-6})(6x10^{-6})}{12}$

• $r = \sqrt{K \frac{(2x10^{-6})(6x10^{-6})}{12}} = \sqrt{(9x10^9) \frac{(2x10^{-6})(6x10^{-6})}{12}} = 9.49x10^{-2} m$

• $q_1 = +2 \mu C +(-4 \mu C) = -2 \mu C = -2x10^{-6} C$

• $q_2 = +6 \mu C +(-4 \mu C) = +2 \mu C = +2x10^{-6} C$

• $F=K \frac{q<em>1 q</em>2}{r^2} = (9x10^9) \frac{(2x10^{-6})(2x10^{-6})}{(9.49x10^{-2})^2} = 4.0 N$

Test Charge ($q_0$)

• A charge with a magnitude small enough to have a negligible effect on the electric field around it.

• A unit positive charge having a magnitude of 1 C.

• The test charge is very small so that the electric field produced by it has a negligible effect.

Electric Field (E)

• The space surrounding an electric charge ‘q’ in which another charge ‘$q_0$’ experiences electrostatic force of attraction or repulsion is called the electric field of the charge ‘q’.

• The direction of the electric field ‘E’ is the same as that of the direction of the force ‘F’.

• The electric field strength EA > EB.

• Closely spaced electric field lines indicate a high intensity of the field, while widely spaced field lines indicate a low intensity of the field.

Intensity or Strength of Electric Field

• It is the ratio of the force acting on the test charge placed at that point to the magnitude of the test charge.

Electric Field due to a Point Charge

• If an isolated point charge '+q' is placed at 'O', and a test charge '+$q_0$' is brought to a point 'P' at a distance 'r' from 'O' and experiences a force 'F', then 'E' will be:

Formula

• $E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$ or $E = K \frac{q}{r^2}$

Vector form of electric field

• $\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^3} \vec{r}$

• $\vec{E} = K \frac{q}{r^3} \vec{r}$

Variation of electric field with distance

• $E \propto \frac{1}{r^2}$

Electric field in terms of surface charge density.

• Surface charge density (σ): Charge per unit area.

• $\sigma = \frac{Q}{A}$

• SI unit of surface charge density: C/m²

Physical significance of electric field

• Two charges kept at a distance exert force on each other, which is termed as action at a distance.

• It is a vector field.

• The electric field concept is also essential for understanding a self-propagating electromagnetic wave such as light.

• The electric field concept gives us a way to describe how starlight travels through vast distances of empty space to reach our eyes.

Importance of electric field in daily life?

Electric field lines and their properties

• The field lines never intersect each other.

• The field lines are perpendicular to the surface of the charge.

• The magnitude of charge and the number of field lines are proportional to each other.

• The electric field is radially outward from positive point charge and radially towards a negative point charge. In other words, electric field lines originate from a positive charge and terminate on the negative charge.

• Electric field lines never form a closed loop because the electric field is conservative.

• In the case of a single charge, electric field lines end at infinity.

• Electric field is strong in the regions where lines of force are closer, and it is weaker where the lines of force are farther apart.

• The lines of force contract longitudinally (lengthwise). This property leads to explaining attraction between two unlike charges.

• The lines of force exert a lateral pressure on each other. This property leads to explaining the repulsion between two like charges.

• The tangent at any point on the electric lines of force gives the direction of the electric field at that point.

Diagrammatic Representation electric field lines

• (i) Isolated point charge

• (ii) Two unlike point charges (Positive and negative)

• (Positive charge)

  • Field lines radially outward

• (ii)

  • Lines of force have a tendency to contract length/longitudinal between two opposite charges and hence they attract each other.

• (Negative charge)

  • Field lines radially inward.

• (iv) Two like point charges (Both positive)

• (v) Two like point charges (Both negative)

  • Lines of force expand laterally between similar charges and hence they repel each other.

Questions

• Q. Draw field lines to represent isolated positive charges.

• Q. Draw field lines to represent isolated negative charges.

• Q. Draw field lines to represent two equal and opposite charges.

• Q. Draw field lines to represent two equal positive charges.

• Q. Draw field lines to represent two equal negative charges.

Electric flux

• The electric flux is a measure of the number of electric field lines passing through a given area.

• It is a scalar quantity.

Electric flux depends upon the following:

• i. Magnitude of electric field (E)

• ii. Area (A)

• iii. Angle between the area vector and direction of the electric field (θ).

• S.I unit: newton meter squared per coulomb

Similarities between electric and gravitational field/force

Limitations of Coulomb's law

1. Coulomb's law holds good for stationary point charges as moving charges produce magnetic field.

2. It is difficult to apply Coulomb's law when the charges are in arbitrary shape. We cannot determine the value of r in this case.

3. Coulomb's law is not valid for all distances. E.g., nuclear distance. The separation between the charges must be greater than nuclear size. If it is less, the nuclear force dominates over the electrostatic force.

Differences between electric and gravitational field/force

Special cases

• (a) θ = 0°

  • Φ = E A cos 0°

  • Φ = E A x 1

  • Φ = EA (Maximum electric flux)

• (b) θ = 30°

  • Φ = E A cos 30°

  • Φ = E A x $\frac{\sqrt{3}}{2}$

• (c) θ = 90°

  • Φ = E A cos 90°

  • Φ = = E A x 0

  • Φ = 0 (Zero electric flux)

• (d) θ = 180°

  • Φ = E A cos 180°

  • Φ = E A x (-1)

  • Φ = -EA (negative electric flux)

Therefore, electric flux can be:

• (i) maximum,

• (ii) zero, and

• (iii) negative.

  • If (i) θ = 0°, flux is maximum,

  • (ii) θ < 90° (acute angle), the flux is positive, and

  • (iii) θ > 90° (obtuse angle), the flux is negative.

Calculations

• The electric field strength just outside a charged conductor is $\sigma/\epsilon_0$

• The electric field strength just inside a charged conductor is 0.

Electric field strength between two charged parallel plates

• If a negative charge (electron) moves from the positive plate to the negative plate, work is done by the charge to overcome the force of repulsion.

• This work done is stored in the form of potential energy in the negative charge.

• Similarly, if a positive charge (proton) moves from the negative plate to the positive plate, work is done by the charge to overcome the force of repulsion.

• Thus, due to potential energy, there is a potential difference between the plates.

Therefore, the electric field strength between the charged parallel plates is the ratio of the potential difference applied between the two parallel plates and the distance between them.

• Conclusions from the expression:

  • i. When the voltage between the plates is increased, the electric field strength will also be increased and vice-versa.

  • ii. When the plates are moved further apart, then the electric field strength will decrease and vice-versa.

Charged particle moving in a uniform electric field.

• If a charged particle, say an electron of mass m, is subjected to a uniform electric field along the X-axis with an initial velocity, v, it will be accelerated by the electric field, which will be directed along the Y-axis following a parabolic path.

• Force of a moving charge along the y-axis = Force due to an electric field

Along the x-axis:

• $u_x = v$

• $S_x = x$

• $a_x = 0$

• Time taken to cover a horizontal direction x is, $t = \frac{x}{v}$

Vertical direction along the y-axis:

• $u_y = 0$

• $S_y = y$

• $a_y = \frac{qE}{m}$

• Using the equation of motion: $S<em>y = u</em>y t + \frac{1}{2} a_y t^2$

• $y = 0 + \frac{1}{2} (\frac{qE}{m}) (\frac{x}{v})^2$

• $y = \frac{qE}{2mv^2} x^2$

• Time to cover horizontal and vertical is the same.

If the charge is an electron, then:

• Q. Prove that the electron follows a parabolic path in a uniform electric field.

• Q. Explain how variables such as initial velocity, charge of particle and mass of particle affect the trajectory of a charged particle in a uniform electric field.

  • Mass – the greater the mass, the smaller the deflection and vice versa.

  • Charge – the greater the magnitude of the charge of the particle, the greater the deflection and vice versa.

  • Initial velocity – the greater the speed of the particle, the smaller the deflection and vice versa.

Question

• When a piece of polythene is rubbed with wool, a charge of $-2 \times 10^{-7} C$ is developed on the polythene. What is the mass transferred on the polythene?

• Total electrons transferred:

  • $Q = ne \Rightarrow n = \frac{Q}{e} = \frac{2 \times 10^{-7} C}{1.6 \times 10^{-19} C} = 1.25 \times 10^{12}$ electrons.

• Total mass transferred on the polythene, $m = n \times$ mass of an electron

• $m = 1.25 \times 10^{12} \times 3.1 \times 10^{-31} kg = 3.875 \times 10^{-19} kg$

  • (mass of an electron = $3.1 \times 10^{-31} kg$)

Question

• The force of attraction between two point charges at a distance r apart is F.

  • (a) What should be the distance apart in the same medium so that the force becomes F/3?

  • (b) What if the force becomes 3F?

Dielectric Question

1. Name the dielectrics above. Ans: Water

2. What is the dielectric constant value above? Ans: 80

3. Why is the dielectric constant also called relative permittivity? Force is calculated with reference to vacuum.

4. How does the force between the charges change in water compared to air? Ans: Force reduces as water has a higher permittivity value

5. If the dielectric constant of a material is 1, what does it tell about the material? Ans: The material behaves like a vacuum or free space; it does not reduce the electric field.

6. In which case will the force between charges be the strongest? Air? Water? Glass? Ans: In air because it has a lower dielectric constant than water and glass.

Question

• Two charged particles $q<em>1 = +8 \mu C$ and $q</em>2 = +2 \mu C$ lie along the x-axis at a distance of 6.00 m apart. Locate the point where the resultant electric field is zero.

• $E<em>1 = K \frac{q</em>1}{x^2}$

• $E<em>2 = K \frac{q</em>2}{(6-x)^2}$

• $E<em>1 - E</em>2 = 0$

• $E<em>1 = E</em>2$

• $K \frac{8 \times 10^{-6}}{x^2} = K \frac{2 \times 10^{-6}}{(6-x)^2}$

• $\frac{8}{x^2} = \frac{2}{(6-x)^2}$

• $4(6-x)^2 = x^2$

• $2(6-x) = x$

• $12 - 2x = x$

• $3x = 12$

• $x = \frac{12}{3} = 4 m$

• Therefore, 4m from 8μC or 2m from 2μC.

Question

• A comb run through one's dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day?

• When the comb runs through dry hair, it gets charged by friction. The molecules in the paper get polarized by the charged comb, resulting in a net force of attraction.

• If the hair is wet, or if it is a rainy day, the friction between the hair and the comb reduces. The comb does not get charged, and thus, it will not attract small bits of paper.

Question

• Ordinary rubber is an insulator. But the special rubber tires of aircraft are made slightly conducting. Why is this necessary?

• During landing, the tires of aircraft may get highly charged due to friction between tires and the air strip. If the tires are made slightly conducting, they will lose the charge to the earth; otherwise, too much static electricity accumulated may produce a spark and result in fire.

Question

• Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?

• A moving vehicle gets charged due to friction. The inflammable material may catch fire due to the spark produced by the charged vehicle. When a metallic rope is used, the charge developed on the vehicle is transferred to the ground, and so the fire is prevented.

Question

• Two equal balls having equal positive charge q C are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?

Question

• The dielectric constant of water is 80. What is its permittivity?

Question

• If charges that are present in the air are placed in water, describe the force between two charges.