PHYSICS As LEVEL(FORM FIVE) NOTES – STATIC ELECTRICITY

STATIC ELECTRICITY

It can be shown that there are two kinds of charges by rubbing a glass rod with silk and hanging it from a long silk thread. If a second glass rod is rubbed with silk and held near the rubbed end of the first, the rods will repel each other.

On the other hand, a hard rubber rod rubbed with fur will attract the glass rod rubbed with silk. The modern view of bulk matter is that in its normal (i.e., neutral) state, it contains equal amounts of positive and negative charges.

If two bodies like glass and silk are rubbed together, a small amount of charge is transferred from one to the other, upsetting the electric neutrality of each. In this case, the glass would become positive and the silk negative.

FORCE BETWEEN TWO CHARGES OR ANY TWO CHARGED BODIES

Coulomb found that a force exists between two electrically charged bodies and that this force and the distance between the charged bodies obey the inverse square law. If r is the distance between the charged bodies and F is the force of attraction between these charged bodies, then

F ∝ 1/r² …………………………… (i)

Equation (i) above is known as the inverse square law.

It has been found that if Q₁ and Q₂ are two charges, then the force of attraction between them is given by

F ∝ Q₁Q₂ …………………………… (ii)

Combining eq (i) and (ii),

The permittivity of air at normal pressure is only about 1.005 times that of vacuum (ε₀). For most purposes, therefore, we may assume the value of ε₀ for the permittivity of air.

1. Calculate the value of two charges if they repel one another with a force of 0.1 N when situated 50 cm apart in a vacuum.
2. What would be the size of the charges if they were situated in an insulating liquid whose permittivity was 10 times that of vacuum?

Solution

F = 0.1 N

r = 50 cm = 0.5 m

ε₀ = 8.854 × 10^-12 F/m

Q₁ = Q₂ = Q (they repel)

Formula:

Q² = F × 4πε₀ r²

Q = [0.1 × 4 × 3.14 × 8.854 × 10^-12 × (0.5)²]^1/2

Q = 1.67 × 10^-6 C

(b) Given:

F = 0.1 N

Q₁ = Q₂ = Q

ε = 10 ε₀

r = 50 cm = 0.5 m

Formula:

Q² = F × 4πε r²

Q = [0.1 × 4 × 3.14 × 10 × 8.854 × 10^-12 × (0.5)²]^1/2

Q = 5.27 × 10^-6 C

The distance between the electrical proton in the hydrogen atom is about 5.3 × 10^-11 m.

Mass of electron, m_e = 9.11 × 10^-31 kg

Mass of proton, m_p = 1.67 × 10^-27 kg

Charge, e = 1.6 × 10^-19 C

Gravitational constant, G = 6.67 × 10^-11 N·m²/kg²

Gravitational force between particles:

F_g ∝ (m_e m_p)/r²

F_g = 3.7 × 10^-47 N

Electric force between particles:

F_e = 8.1 × 10^-8 N

ELECTRIC FIELD INTENSITY (E)

Electric field intensity is defined as the region in which an electric force is experienced. So, an electric field intensity E at any point in an electrostatic field is defined as the force per unit charge which it exerts on a positive charge.

If Q is a small test charge placed at a point, then

The SI unit of E is N/C.

QUESTION

Find the magnitude of an electric field strength such that an electron placed in the field would experience an electrical force equal to its weight. Mass of electron = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, g = 9.8 m/s².

Consider a test charge Q₀ in vacuum which is placed a distance “r” from a point charge Q.

By Coulomb’s Law, the magnitude of the force acting on Q₀ is given by

By putting eq (i) in eq (ii),

Note:

1. E is a vector quantity.
2. From ε₀,
3. The direction of ε₀ is radial line from Q point outwards if Q is positive and inwards if Q is negative.

LINES OF FORCE

There is a relationship between imaginary lines of force and the electric field strength as follows:

(i) The tangent to a line of force at any point gives the direction of E at that point.

The lines of force are drawn so that the number of force-permitting cross-sectional areas is proportional to the magnitude of E. Where the lines are close together, E is large; where they are far apart, E is small.

ELECTRIC FLUX (Ø)

The electric flux Ø through an area perpendicular to the total lines of force is the product of the electric field intensity E and the area A where E is the electric intensity at that place.

Consider a sphere of radius r drawn in space concentric with a point charge.

Total flux through the sphere is given by

Ø = E × Area of the sphere

= E × 4πr²

If the charge is placed in any other medium apart from air or vacuum, then

The above equation shows that the total flux crossing any point on the drawn sphere concentrically outside the point charge is constant.

1. Outside the charged sphere

The flux across a spherical surface of radius r concentric with a small sphere carrying charge Q is given by

This result shows that outside a charged sphere, the field behaves as if all charges on the sphere are concentrated at the centre.

2. Inside a charged empty sphere

Inside the empty charged sphere there are no charges, so the electric field strength E = 0; therefore, since Q = 0, the value of E inside the sphere is also 0.

ELECTRIC INTENSITY OUTSIDE THE CHARGED PLATES

Consider a charged plane conductor S with a surface charge density of 6 μC/m². Let the plane surface 1 as shown above be drawn outside the S which is parallel to S and has the area A m².

The intensity of the field must be perpendicular to the surface, and the charges will produce this field in the projection of the area P on the surface S, i.e., those within the shaded region A.

Question

A particle of mass M and charge q is placed at rest in a uniform electric field (see the figure below) and released. Describe its motion.

The motion resembles that of a falling body in the earth’s gravitational field. The constant acceleration is given by

The equation of uniform acceleration applies, therefore with

The vertical distance moved by a particle with initial velocity

By putting eqn (1) in (3) we get

From the third equation of motion we have

Putting eqn (1) into eqn (5)

The kinetic energy attained at the moving distance y is formed from:

Substitute eqn (6) into (7) we get

ELECTRIC POTENTIAL (V)

The electric field around a charged body can be described not only by a vector electric field strength E but also by a scalar quantity, i.e., the electric potential, V.

To find the electric potential difference between two points A and B in an electric field, we move a test charge q from A to B and measure the work W_AB that must be done by the agent moving the charge.

Electric potential difference, V can be expressed as

The unit of potential difference is joule per coulomb (J/C), also called volts.

1 J/C = 1 Volt

If point A is chosen to be very far (say at infinity), then the electric potential at infinity is arbitrarily taken as zero.

Therefore, putting V_A = 0 in equation (i) and dropping the subscripts, we get

Definition:

The electric potential at a point is the work done by the force in taking the unit charge from infinity to that point.

CALCULATION OF WORK DONE

Consider a positive charge Q at a distance R_a as indicated in the figure. The work done in taking the charge from A to B is equal to the work done in taking the same distance from B to A. If Q₀ is moved by the force from A to B, then the force acting on it is

If the charge has moved a distance dx, the work done is

dw = F dx

Hence total work done in taking Q₀ from A to B is

We get

W_AB = Q₀ (V_B – V_A)

The potential difference between A and B is given by

Substitute eq-(ii) into (iii)

We get

Therefore

If r_B is very large, i.e., B is at infinity, the potential at A is given by

So in general, the potential V at a distance r from the point charge Q is given by:

POTENTIAL DUE TO SEVERAL CHARGES

The potential at any point due to a group of charges is found by:

1. Calculating the potential V_n due to each charge as if other charges are not present.
2. Adding the quantities so obtained.

Where Q is the value of the charge and r is the distance of this charge from the point in question.

Question

Calculate the potential at the centre of the square shown below:

Eight charges having the values shown in the figure below are arranged systematically on the circle of radius 0.4 m in air. Calculate the potential at the centre O.

RELATIONSHIP BETWEEN ELECTRIC INTENSITY (E) AND ELECTRIC POTENTIAL V:

Consider two points A and B at distances x and x + dx from O respectively. V_A and V_B are the potentials at A and B respectively. A and B are very close so that the electric intensity E is instant. Hence the potential difference between A and B is

V_AB = V_A – V_B

TYPES OF CAPACITOR

(i) VARIABLE AIR CAPACITOR

Variable (capacitance) capacitor is one in which the effective area of the plates can be adjusted. The capacitance of a variable capacitor can be varied as you wish but within certain limits.

These are widely used in the tuning circuits of radio receivers. They are constructed of a number of affixed parallel metal plates. The plates connected together constitute one plate of the capacitor. The second set of movable plates also connected together form the other plate.

By rotating the plates on which the movable plates are mounted, the second set may be caused to interleave the first to a lesser or greater extent.

The effective area of the capacitor is that of the interleaved portion of the plates only.

The plates of the capacitor may be made of brass or aluminum. The dielectric may be oil, air, or mica.

(ii) A MULTIPLE CAPACITOR MICA DIELECTRIC

The plates of this type of capacitor are made of tin foil. The capacitance of the capacitor is n times the capacitance between two successive plates, where n is the number of dielectrics between the plates.

(iii) PAPER CAPACITOR

The paper capacitor has a dielectric of paper impregnated with paraffin wax or oil. Unlike the mica capacitor, the paper can be rolled and stacked into a cylinder of relatively small volume. Nowadays, the paper has been replaced by a thin layer of polystyrene.

(iv) ELECTROLYTIC CAPACITOR

They are produced by passing a direct current between two sheets of aluminum foil with a suitable electrolyte of lightly conducting liquid between them. A very thin film of aluminum oxide is then formed on the oxide plate which is on the positive side of the DC supply.

This film is an insulator. It forms the dielectric between two plates, the electrolyte being a good conductor.

Since the dielectric thickness d is very small and C ∝ 1/d, the capacitance value can be very high.

ARRANGEMENT OF CAPACITOR

1) Parallel arrangement of capacitors

All the left-hand plates are connected together and all the right-hand plates are connected together. In the case of parallel arrangement of capacitors, when a cell is connected across these capacitors in parallel, they have the same potential difference (V).

So,

1. Q₁ = C₁ V
2. Q₂ = C₂ V
3. Q₃ = C₃ V

Let the total charge be Q, then Q = Q₁ + Q₂ + Q₃ ……… (4)

2) SERIES ARRANGEMENT OF CAPACITORS

When the right-hand plate of one capacitor is connected to the left-hand plate of the next and so on, these capacitors are said to be connected in series.

When the cell is connected across the end of the system, a charge is transferred from the plates. This charge induces a charge on the adjacent plate. This process is repeated with other plates.

Question

Two capacitors of capacitance C₁ = 2 μF and C₂ = 8 μF are connected in series and the resulting combination is connected across 300 volts. Calculate the charge and potential difference.

Solution

C₁ = 2 × 10^-6 F

V₁ = 200 V

C₂ = 8 × 10^-6 F

V₂ = 100 V

Energy stored in capacitors:

E₁ = ½ C₁ V₁² = ½ × 3 × 10^-6 × (200)² = 6 × 10^-2 Joules

E₂ = ½ C₂ V₂² = ½ × 2 × 10^-6 × (100)⁴ = 10^-2 Joules

Total energy stored = 5.4 × 10^-2 Joules

ENERGY STORED IN A CAPACITOR

Consider a capacitor of capacitance C charged to a potential difference V. Let a small charge dQ be transferred from the negative plate to the positive plate. Then the work done in moving a charge dQ will be

dw = V dQ, but V = Q/C

Suppose a capacitor is at first discharged and then charged until the final charge on the plate is Q. The work done in charging it is given by

In general, if C is the capacitance of a capacitor carrying charge Q at potential difference V, then

W = ½ QV

Also, C = Q/V

Putting into the above, we get

W = ½ CV²

DISCHARGE IN C-R CIRCUIT

Consider a capacitor initially charged to a potential difference V₀ so that its charge is Q = CV₀.

At time t, after the discharge through R has begun, the current I flowing is given by

I = V/R, where V is the potential difference across C.

Now, V = Q/C and I = dQ/dt

The minus sign shows that Q decreases with increasing time. From the equations, we have

Integrating the above expression gives

Let U = Q₀ – Q, du = -dQ

ln(Q₀ – Q) – ln Q = ln(Q₀/Q)

From equation (v), Q decreases exponentially with time.

Since the potential difference V across C is proportional to Q, then V = V₀ e^-t/CR. Also, since the current I is proportional to V, then I = I₀ e^-t/CR, where I₀ is the initial current value.

The time constant (τ) of the discharge circuit is defined as CR seconds, where C is in Farads and R is in Ohms.

Example: A resistor of resistance R = 10 Ω is connected in series with a capacitor of capacitance 1 μF. Find the time constant and half-life.

Solution:

C = 1 × 10^-6 F

R = 10 Ω

Time constant τ = CR = 10 × 10^-6 = 10 seconds

Half-life T_1/2 = 0.693 × τ = 6.93 seconds

CHARGING OF CAPACITOR THROUGH A RESISTOR

Consider charging a capacitor with capacitance C through a resistor R in series.

Suppose the supplied battery has an emf E and negligible internal resistance.

Initially, there is no charge on the capacitor, so no potential difference across it. After connecting the battery, the potential difference across R = E, and the initial current flow I₀ = E/R.

Suppose I is the current flowing after time t, and V_C is the potential difference across C.

Equation:

CR dV_C/dt + V_C = CE

CE is the final charge on C when no further current flows.

Integrating the above expression gives

CAPACITANCE

It can be shown by sending a positive or negative charge close to a charged body that the potential difference between the system of charged bodies is proportional to the charge.

If Q is the charge and V is the potential difference, then

Q ∝ V, i.e., Q = CV ………………… (i)

Where the constant C is known as capacitance.

The capacitance of a system of bodies is the charge necessary to raise the potential by a unit, i.e.,

C = Q/V

Capacitance is measured in Farads, but other units like μF and pF are also used.

The symbol of a capacitor is

THE DERIVATION OF CAPACITANCE IN PARALLEL PLATES CAPACITOR

Consider two parallel plates capacitor with plates carrying charge Q.

The plates have electric flux Ø = E × A ……………(1)

But also Ø = Q/ε₀ …………………………… (2)

From equations (1) and (2),

E A = Q/ε₀ or E = Q/(A ε₀) ………………………..(3)

The work required to take a test charge from one plate to another is

W = Force × distance = E Q d ……………………… (4)

But also work done

W = Q V ………………………………… (5)

Q V = ε₀ Q d

V = ε₀ d / A …………………………… (6)

Put equation (3) into (6), we have

V = Q d / (A ε₀) …………………….. (7)

But capacitance, C = Q / V ……………………… (8)

Substitute eqn (7) into (8), we get

C = A ε₀ / d

If the distance between plates is filled with air or vacuum, the equation above becomes

C = A ε₀ / d

FACTORS WHICH DETERMINE CAPACITANCE

1. Distance between the plates
   From the equation V = Q d / (A ε₀), it follows that if the plates are farther apart, the potential difference will increase and hence capacitance C = Q / V decreases. Therefore, the capacitance decreases when the separation of the plates increases.
2. Dielectric
   When a dielectric (e.g., sheet of glass or ebonite) is placed between the plates, the potential difference between the plates decreases. Hence, the capacitance increases. Therefore, when the dielectric constant ε_r increases, the capacitance also increases.
3. Area of plates
   From V = Q d / (A ε₀), it follows that as the area of the plates increases, the potential difference decreases, so the capacitance C = Q / V must increase.

DIELECTRIC CONSTANT (ε_r)

Dielectric constant is also known as relative permittivity.

The ratio of the capacitance with dielectric to the one without dielectric between the plates is called the dielectric constant or relative permittivity of the material used.

Consider the case of a parallel plates capacitor with capacitance with dielectric C = A ε / d and capacitance without dielectric C₀ = A ε₀ / d.

Hence,

ε_r = ε / ε₀ ………………… (i)

From equation (i), it follows that the capacitance of a capacitor can also be given as

A capacitor which leads to capacitance is a device for storing charges. Essentially all capacitors have metal plates separated by an insulator called dielectric. In some capacitors, dielectrics used are oil, air, polyethylene, etc.

MEASUREMENT OF SURFACE TENSION BY CAPILLARY TUBE

Suppose a clean glass capillary tube is dipped into water; water level rises and the angle of contact is zero. The figure below shows a section of the meniscus M at B which is hemispherical.

Glass AB is tangent to the liquid at its meniscus.

The surface tension force acts along the boundaries of the liquid and the air vertically downwards on the glass.

According to action and reaction, the glass exerts an upward force on the liquid meniscus. If r is the radius of the capillary tube, the length of water in contact with glass is 2πr.

Upward force on liquid (I) = 2πr γ

This force supports the weight of the liquid column of height h above the outside level.

Volume of the liquid = π r² h

Mass of the liquid = ρ π r² h (where ρ is density)

Weight of liquid = ρ g π r² h

From (i), we have

2 π r γ = ρ g π r² h

Therefore, the angle of contact is assumed equal to zero. The weight of the small amount of liquid above the meniscus has been neglected.

CAPILLARY RISE AND FALL BY PRESSURE METHOD

In figure (I) below, the liquid rises up the tube to a height h, so that the pressure P₁ has

a less pressure than P₂, and

If the capillary tube is dipped into water, the angle of contact is practically zero. This is the atmospheric pressure and P₁ is the pressure in the liquid. We have

P₂ – P₁ = 2 γ cos θ / r

If H is the atmospheric pressure, h is the height of the liquid in the tube, and ρ is its density, then

P₂ = H and P₁ = H – ρ g h

Therefore,

h = 2 γ cos θ / (ρ g r)

γ = 7.5 × 10^-2 N/m

Questions

1. Water rises to a height of 5 cm in a certain capillary tube. In the same tube, mercury is depressed by 1.71 cm. Compare the surface tension of water and mercury. Specific gravity of mercury is 13.6, angle of contact for water is zero and that of mercury is 135°.
2. A liquid of surface tension γ is used to form a film between a horizontal rod of length L and another shorter rod of mass m supported by two light inextensible strings of equal length joining adjacent ends of each rod. The film fills the vertical plane within rods and strings. What is the shape of each string?
3. A soap bubble in vacuum has a radius of 3 cm and another soap bubble in vacuum has a radius of 6 cm. If the two bubbles coalesce under isothermal conditions, calculate the radius of the bubble formed under isothermal conditions assuming γ is constant.