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Unit 9: Electrostatics and Current Electricity — Short Questions

11th Class Physics · Unit 9: Electrostatics and Current Electricity

Exercise Short Questions

9.1.How does a moving conductor like an aeroplane acquire charge as it flies through the air? Describe briefly.

An airplane can become charged due to friction with air molecules, resulting in a transfer of electrons and a buildup of static electricity. The airplane often becomes charged. It can become either:
Positively charged (losing electrons) or
negatively charged (gaining electrons) the specific charge depends on factors like air speed, humidity, and the materials involved.

9.2.Define electric intensity and electric potential.

Electric Intensity (Electric Field Intensity)
The electric intensity, or electric field intensity, at a point is the force per unit charge experienced by a small test charge placed at that point.
Mathematically:
Electric Intensity (E): E = F / q
Where:

Electric Potential
Electric potential, is the potential difference per unit charge between two points in an electric field. It's the work done in moving a unit charge from one point to another.
Mathematically:
Electric Potential (V): V = W / q
Where:
V = electric potential (V)
W = work done in moving a charge (J)
Electric potential is a scalar quantity and is measured in volts (V).

9.3.A battery is rated at 100 Ah (ampere-hour). How much charge can this battery supply?

The value of charge supplied by battery is given by formula
Q = It
I = 100A
t = 1h = 3600s
Q = 100 × 3600 = 3.6 × 10⁵ C.

9.4.Is electron-volt a unit of potential difference or energy? Explain.

An electron volt (eV) is a unit of energy, not potential difference.
1 electron volt is the energy gained or lost by an electron when it moves through a potential difference of 1 volt.
Mathematically:
1 eV = 1.602 × 10⁻¹⁹ J (joules)

9.5.A copper wire of length L has resistance R. It is stretched to double its length. What will be the of its resistance of the new length of wire?

When the copper wire is stretched to double its length, its volume remains constant. As a result, the cross-sectional area decreases to half.
Resistance (R) is given by:
R = ρ × L / A
When length (L) doubles and area (A) halves:
New length (L') = 2L
New area (A') = A / 2
New resistance (R'):
R' = ρ × (2L) / (A / 2)
R' = ρ × 2L × 2 / A
R' = 4 × ρ × L / A
R' = 4R
The new resistance will be four times the original resistance.

9.6.Why does the resistance of a conductor rise with increase in temperature?

The resistance offered by a conductor to the flow of electric current is due to collisions of free electrons with the atoms. As the temperature of the conductor rises the amplitude of vibration of the atoms increases and the chances of their collision with the free electrons also increases. In this way resistance of conductor increases.
The increases in resistance is in accordance with relation
RT = Ro (1 + αΔT)

9.7.Is the filament resistance lower or higher in a 500W-220V light bulb than in a 100 W-220V bulb?

To determine which light bulb has a higher filament resistance, we use the formula: Resistance R = V² / P
For the 100W-220V bulb:
R = V² / P = 220² / 100 = 484 ohms
For the 5000W-220V bulb:
R = V² / P = 220² / 5000 = 9.68 ohms
The filament resistance is lower in the 5000W-220V light bulb than in the 100W-220V bulb.

9.8.Why does resistance of a thermistor decrease as temperature increases?

The resistance of negative temperature co-efficient (N.T.C) decreases with temperature because it is made from semiconductor materials as temperature of sami conductor increases more free electrons and holes (charge carriers) are generated.
These increased number of charge carriers enhance conductivity, thus, decreases resistance.

9.9.Which materials can be used to construct Faraday's cage and why?

A Faraday cage can be made from conductive materials like:
i. Copper
ii. Aluminum
iii. Steel

The conductive materials allow electric charge to move freely across the surface of cage when exposed to electric field.
This redistribution of charge creates an opposite electric charge cancels out the external field inside the cage.

SLO Based Additional Short Questions + Past papers Short Questions of Punjab Boards

Electric force

Q1.What is electric force?

It is force between the two charge particles which may be force of attraction or repulsion. It depends upon the quantity of charge, the distance and the nature of medium. This force exists between the atoms molecules and ions of a substance.

Coulomb's law

Q2.State Coulomb's law. Write its mathematical form.

The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. If there are two charges q₁ and q₂ separated by a distance r, then electrostatic force F between charges, according to Coulomb's law is
F ∝ q₁ q₂
F ∝ 1/r²
Combining
F ∝ q₁q₂/r²
F = kq₁q₂/r²

Coulomb's constant

Q3.Find the nature of k in Coulomb's law.

The Coulomb's force is
F = kq₁q₂/r²
Where k is Coulomb's constant. The value of k depends upon:
1. System of units used
2. Nature of the medium In vacuum or air
k = 1/(4πε₀)
Where ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is electrical constant, known as permittivity of free space.
k = 9 × 10⁹ Nm²C⁻²

Effect of medium on coulomb's force

Q4.What is the effect of medium on Coulomb's force?

It has been found experimentally that if an insulator is used as a medium in place of air or vacuum, the force between the electric charges is reduced by a factor εᵣ, known as relative permittivity. The force between two charges in the presence of medium is
F' = 1/(4πε₀εᵣ) × q₁q₂/r²
F' = 1/εᵣ × (1/(4πε₀) × q₁q₂/r²)
But
F = 1/(4πε₀) × q₁q₂/r²
So,
F' = F/εᵣ
The value of εᵣ is always greater than 1.
So F' < F

Electric intensity

Q5.Calculate the electric field intensity due to a point charge.

Consider a point charge q placed at point O. A test charge q₀ is placed at P in an electric field at a distance r
According to Coulomb's Law, the force experienced by a point charge q₀ at P due to charge q is,
F = 1/(4πε₀) × q q₀/r²
As electric intensity at P is
E = F/q₀
So
E = 1/(4πε₀) × q/r²

Application of a hollow charged metal sphere

Q6.The interior of a hollow charged metal sphere is a field free region. Where this fact is applied?

Any electrical circuit placed inside this metal enclosure is 'shielded' from stray electric field.

Gaussian surface

Q7.What is a Gaussian surface? How flux is calculated through a Gaussian surface?

An imaginary closed surface passing through the point at which electric intensity is to be calculated is known as Gaussian surface.
Flux through the Gaussian surface is calculated by Gauss's law which is written as
φₑ = q/ε₀

Potential gradient

Q8.What do you mean by potential gradient?

The quantity ΔV/Δr gives the maximum value of rate of change of potential with respect to distance because the charge is moved along the path along which the distance between the points is minimum. It is known as potential gradient.
Electric intensity can also be called as negative gradient of potential. The negative sign shows that the direction of E is along the decreasing potential.
E = - ΔV/Δr

Relation between bolt and metre

Q9.Prove 1 volt / 1 metre = 1 newton / 1 coulomb

1 volt / 1 metre = 1 joule/coulomb / 1 metre
1 volt / 1 metre = 1 newton × 1metre / 1 metre × 1 coulomb
1 volt / 1 metre = 1 newton / 1 coulomb

Electron volt

Q10.What is an electron – volt? Show that 1ev = 1.6 10⁻¹⁹ J.

It is defined as the energy gained or lost by an electron when it is moved between two points with a potential difference of one volt
as Δ (K.E) = q ΔV
If ΔV = 1 volt "and" q = e = 1.6 × 10⁻¹⁹ C
Then
Δ (K.E.) = (1.6 × 10⁻¹⁹ C) (1 Volt)
1eV = 1.6 × 10⁻¹⁹ joule

Current electricity

Q11.What is current electricity?

The branch of Physics which deals with the charges in motion is called electrodynamics or current electricity.

Electric current

Q12.Define electric current and its units.

The amount of charge per unit time passing through any cross-section of a conductor is called electric current.
I = Q/t. When one coulomb of charge passes through any cross section of a conductor in one second, the current is said to be one ampere. i.e. 1 A = 1Cs⁻¹

Conventional current

Q13.What is conventional current?

The current due to the motion of positive charges, that flows from positive terminal of battery to its negative terminal is called conventional current. In case of analyzing the electric circuit, we use the direction of the conventional current.

Drift velocity

Q14.What is drift velocity?

The uniform velocity of free electrons inside the metallic conductor in opposite direction of E, when steady current flows through the conductor, is called drift velocity. Its value is of the order of 10⁻³ ms⁻¹.

Resistivity

Q15.What is resistivity or specific resistance?

Resistivity of a substance is defined as resistance of a meter cube of the substance. It is measured in ohm-meter.

Conductivity

Q16.What is conductance and conductivity?

Conductance is defined as reciprocal of resistance.
Conductance = 1/Resistance
Unit of conductance is mho (or siemen) or ohm⁻¹. The reciprocal of resistivity is called conductivity.
Conductivity = 1/Resistance

Effect of temperature on resistance

Q17.What is the effect of temperature on resistance?

With the increase in temperature, the resistance of a conductor also increases. This is due to the increased collisions between the free electrons and the atoms of the conductor.

Temperature coefficient of resistance

Q18.Define temperature coefficient of resistance.

The fractional change in resistance per Kelvin is known as the temperature coefficient of resistance. Mathematically.
α = (R - Rt) / Rt
(OR) The fractional change in resistivity per Kelvin is known as coefficient of resistivity.
α = (ρt - ρo) / (ρo × t)

Electromotive force

Q19.What is electromotive force?

The EMF of the source is defined as the energy supplied to unit charge by the cell.
(OR) It is defined as the potential difference between the terminals of a battery when no current is flowing through an external circuit or when the circuit is open. Mathematically.
E = Δw/Δq

e.m.f. and potential difference

Q20.What is the difference between electromotive force and terminal potential difference?

The EMF is the "cause" and the potential difference is its "Effect" The EMF is always present even when no current is drawn from the battery or cell, but the potential difference across the conductor is zero when no current flows through it

Kirchhoff's rule

Q21.What are the rules for finding the potential changes when we apply Kirchhoff's 2nd rule?

(i) If a source of e.m.f. is traversed from negative to positive terminal, the potential changes is positive, it is negative in the opposite direction.
(ii) If a resistor is traversed in the direction of current, the change in potential is negative, it is positive in the opposite direction.

Potentiometer

Q22.What is a potentiometer? Give its principle.

A very simple instrument, which can measure and compare potentials without drawing any current from the circuit, is called potentiometer. The principal of potentiometer is as follows:
The potential difference across any length of wire of uniform area of cross-section is directly proportional to length when a constant current flows through it.

Q23.Give uses of potentiometer.

Potentiometer is used to (a) Determine e.m.f. of a cell
(b) Compare the e.m.f. of two cells.
(c) As a potential divider.
(d) To measure accurate value of potential difference.

Constructed Response Questions

9.1.Electric lines offeree never cross each other. Why?

Electric lines of force (electric field lines) never cross each other because:
If they do so, it would imply two different directions of the electric field at the point of intersection, which is not possible. The electric field at any point has a unique direction, tangent to the field line.

9.2.Is E necessarily zero inside a charged rubber balloon if the balloon is spherical? Assume that charge is distributed uniformly over the surface.

According to Gauss's Law, the net electric flux through a closed surface is proportional to the charge enclosed within that surface.
Mathematically:
φ = Q/ε₀ As Q = 0 (inside the spherical balloon)
φ = 0
Also φ = E⃗ · A⃗
0 = E⃗ · A⃗
E⃗ = 0 (∵ A⃗ ≠ 0)

9.3.Electrostatic force is 10³⁸ times stronger than gravitational force. Argue that our galaxy should be almost electrically neutral.

As electrostatic force is 10³⁸ times stronger than gravitational force, even a tiny imbalance in charge would overpower gravity. For a galaxy to maintain its structure, it must be nearly electrically neutral, with positive and negative charges balancing each other out. Otherwise, electrostatic forces would dominate, disrupting the galaxy's stability. Thus, galaxies are expected to be almost electrically neutral.

9.4.An uncharged conducting hollow sphere is placed in the field of a positive charge q. What will be the net flux through the shell?

The net electric flux through a closed surface is given by Gauss's law:
Φₑ = q/ε
As q = zero
So, net electric flux through shall is also zero.

9.5.A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free elections by: i. increasing the potential difference? ii. decreasing the length and the temperature of the wire?

Effect on Drift Velocity i. The drift velocity of free electrons increases by increasing the potential difference. vd ∝ ΔV It is because by increasing the potential difference, the electric field becomes stronger which increases the drift velocity of free electrons in the conductor

ii. We know that the resistance of the conductor is directly proportional to its length. If we decrease the length of the conducting wire, and temperature then its resistance decreases and current increases and thus the drift velocity of free electrons increases.
vd ∝ 1/R

9.6.Why the terminal potential difference of a battery decreases when the current drawn from it is increased?

The terminal potential difference (V) of a battery decreases when current (I) drawn from it increases due to the internal resistance (r) of the battery.
According to Ohm's law: V = ε - Ir
where ε is the electromotive force (e.m.f.) of the battery.
As current increases, the voltage drop (Ir) across the internal resistance increases, reducing the terminal potential difference (V).