Electricity: NMAT Physics Review

Introduction to Electricity

Electricity is one of the core topics in NMAT Physics and plays a crucial role in understanding both classical and modern physical systems. It deals with electric charges, their motion, the forces they exert, and the energy they transfer. In the NMAT, questions on electricity often test conceptual understanding, mathematical problem-solving, and the ability to apply formulas efficiently under time pressure.

This review covers fundamental concepts such as electric charge, electric force, electric field, electric potential, capacitance, current electricity, resistance, and electrical power. Mastery of these topics is essential, as electricity frequently appears in both standalone questions and integrated problems with mechanics or magnetism.

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience an electric force when placed in an electric field. There are two types of electric charges: positive and negative. Like charges repel each other, while unlike charges attract each other.

The SI unit of electric charge is the coulomb (C). The smallest unit of charge is the elementary charge, denoted by e, which has a magnitude of approximately 1.6 × 10^-19 C. Protons carry a positive elementary charge, while electrons carry an equal-magnitude negative charge.

Electric charge is conserved, meaning the total charge in an isolated system remains constant. Charges can be transferred from one object to another through processes such as friction, conduction, and induction.

Coulomb’s Law

Coulomb’s Law describes the electrostatic force between two point charges. The magnitude of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Mathematically, Coulomb’s Law is expressed as:

F = k |q₁ q₂| / r²

where k is Coulomb’s constant (approximately 9 × 10⁹ N·m²/C²), q₁ and q₂ are the charges, and r is the distance between them.

The force acts along the line joining the two charges and can be attractive or repulsive depending on the nature of the charges. NMAT questions often involve calculating forces or comparing force magnitudes when distance or charge values change.

Electric Field

The electric field is a region around a charged object where another charge experiences an electric force. It is defined as the electric force per unit positive test charge placed at a point in the field.

The electric field strength E is given by:

E = F / q

For a point charge, the electric field is expressed as:

E = k q / r²

Electric field is a vector quantity, meaning it has both magnitude and direction. The direction of the electric field is defined as the direction of the force experienced by a positive test charge.

Electric field lines are used as a visual representation of electric fields. They originate from positive charges and terminate on negative charges. The density of field lines indicates the strength of the field.

Electric Potential and Electric Potential Energy

Electric potential is a scalar quantity that represents the electric potential energy per unit charge at a point in an electric field. The SI unit of electric potential is the volt (V), where 1 V = 1 J/C.

Electric potential V is defined as:

V = U / q

where U is electric potential energy and q is the charge.

For a point charge, electric potential is given by:

V = k q / r

Electric potential energy depends on the position of a charge in an electric field. Changes in electric potential energy are important in understanding how charges move and how electrical devices function.

Potential Difference and Voltage

Potential difference, commonly called voltage, is the difference in electric potential between two points. It represents the work done per unit charge in moving a charge from one point to another.

Voltage is what drives electric current in circuits. Batteries and power supplies maintain a potential difference that causes charges to flow.

In NMAT problems, voltage is often related to energy conversion, electric work, and circuit analysis. Understanding the relationship between voltage, current, and resistance is essential.

Capacitance and Capacitors

Capacitance is the ability of a system to store electric charge. A capacitor is a device designed to store charge and electrical energy. It typically consists of two conductors separated by an insulating material called a dielectric.

Capacitance C is defined as:

C = Q / V

where Q is the charge stored and V is the potential difference across the capacitor. The SI unit of capacitance is the farad (F).

For a parallel-plate capacitor, capacitance is given by:

C = ε₀ A / d

where A is the plate area, d is the separation between the plates, and ε₀ is the permittivity of free space.

Capacitors can be connected in series or parallel, and NMAT questions often test the equivalent capacitance and charge distribution in such configurations.

Electric Current

Electric current is the rate of flow of electric charge through a conductor. It is defined as:

I = Q / t

where I is the current, Q is the charge, and t is the time. The SI unit of current is the ampere (A).

In metallic conductors, current is due to the motion of electrons, while in electrolytes and plasmas, it is carried by ions. Conventional current direction is defined as the direction of positive charge flow.

Ohm’s Law

Ohm’s Law relates voltage, current, and resistance in an electrical circuit. It states that the current through a conductor is directly proportional to the voltage across it, provided the temperature remains constant.

Ohm’s Law is expressed as:

V = I R

where V is voltage, I is current, and R is resistance.

This law is fundamental to circuit analysis and is frequently used in NMAT problems involving simple and complex circuits.

Resistance and Resistivity

Resistance is a measure of how much a material opposes the flow of electric current. The SI unit of resistance is the ohm (Ω).

Resistance depends on the material, length, and cross-sectional area of the conductor. It is given by:

R = ρ L / A

where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.

Resistivity is an intrinsic property of a material and varies with temperature. Conductors generally have low resistivity, while insulators have high resistivity.

Electrical Power and Energy

Electrical power is the rate at which electrical energy is converted into other forms of energy, such as heat, light, or mechanical work. Power is given by:

P = V I

Using Ohm’s Law, power can also be expressed as:

P = I² R or P = V² / R

The SI unit of power is the watt (W). Electrical energy consumption is often measured in kilowatt-hours (kWh), especially in household applications.

Series and Parallel Circuits

In a series circuit, components are connected end to end, so the same current flows through each component. The total resistance is the sum of individual resistances.

In a parallel circuit, components are connected across the same voltage source. The voltage across each component is the same, while the total current is the sum of currents through each branch.

NMAT problems often involve analyzing series-parallel combinations, requiring careful application of Ohm’s Law and equivalent resistance concepts.

Common NMAT Electricity Problem-Solving Tips

To excel in electricity questions in the NMAT, it is important to memorize key formulas and understand their physical meaning. Dimensional analysis can help verify answers, and approximation skills are useful for time management.

Drawing circuit diagrams clearly and labeling known quantities can reduce errors. Pay attention to units and conversion factors, especially when dealing with micro, milli, or kilo prefixes.

Conclusion

Electricity is a high-yield topic in NMAT Physics that combines conceptual understanding with quantitative problem-solving. By mastering electric charge, fields, potential, current, resistance, and power, students can confidently tackle a wide range of NMAT questions.

Regular practice, combined with a solid grasp of fundamental principles, is the key to achieving a strong performance in the electricity portion of the NMAT Physics section.

Problem Sets

Set 1: Electric Charge, Coulomb’s Law, and Superposition

1. Two point charges +3.0 μC and −2.0 μC are separated by 0.50 m in air. Find the magnitude of the force between them.
2. If the distance between two point charges is doubled, by what factor does the electrostatic force change?
3. Charges q1 = +4.0 μC and q2 = +1.0 μC are 0.20 m apart. What is the force on q2 due to q1?
4. Three charges lie on a line: qA = +2.0 μC at x = 0 m, qB = −1.0 μC at x = 0.30 m, and qC = +3.0 μC at x = 0.60 m. Find the net force on qB.
5. Two identical charges repel each other with 0.90 N at a distance of 0.10 m. What is the magnitude of each charge?

Set 2: Electric Field and Electric Potential

1. What is the electric field magnitude 0.40 m away from a point charge of +5.0 μC?
2. A positive test charge of 2.0 nC is placed in a uniform electric field of 3.0 × 10⁴ N/C. What is the magnitude of the electric force on it?
3. Find the electric potential 0.25 m away from a point charge of −6.0 μC (take V = 0 at infinity).
4. How much work is required to move a +3.0 nC charge from a point at 120 V to a point at 45 V?
5. Two point charges +2.0 μC and +8.0 μC are located 0.30 m apart. At what point on the line between them is the net electric field zero? (Give distance from the +2.0 μC charge.)

Set 3: Capacitance and Energy Stored

1. A capacitor stores 6.0 μC of charge when connected to a 12 V battery. Find its capacitance.
2. A parallel-plate capacitor has plate area 0.020 m² and separation 1.0 mm. Find its capacitance (use ε₀ = 8.85 × 10⁻¹² F/m).
3. A 4.0 μF capacitor is charged to 9.0 V. How much energy is stored?
4. Two capacitors 3.0 μF and 6.0 μF are connected in series. Find the equivalent capacitance.
5. Two capacitors 2.0 μF and 5.0 μF are connected in parallel to a 10 V source. Find the total charge stored in the combination.

Set 4: Current, Resistance, and Ohm’s Law

1. A potential difference of 24 V is applied across a 6.0 Ω resistor. Find the current.
2. A wire has resistance 3.0 Ω. If the current is 2.5 A, what is the voltage across the wire?
3. A device draws 0.80 A from a 120 V outlet. What is its resistance (assume it obeys Ohm’s Law)?
4. A 12 V battery is connected to two resistors 4.0 Ω and 8.0 Ω in series. Find the current and the voltage drop across each resistor.
5. Two resistors 6.0 Ω and 3.0 Ω are connected in parallel across a 12 V source. Find the total current drawn.

Set 5: Electrical Power, Energy, and Mixed Circuits

1. A 10 Ω resistor carries 1.5 A. Find the power dissipated.
2. A 60 W bulb is connected to a 120 V supply. Find the current it draws.
3. A heater rated 1500 W runs on 220 V. Find its resistance.
4. How much electrical energy (in kWh) is consumed by a 100 W lamp operating for 6.0 hours?
5. A 9 V battery is connected to a circuit where a 2.0 Ω resistor is in series with a parallel combination of 3.0 Ω and 6.0 Ω. Find the total current from the battery.

Answer Keys

Set 1 Answers

1. F = k|q1q2|/r² = (9×10⁹)(3×10⁻⁶)(2×10⁻⁶)/(0.50)²
   = (9×10⁹)(6×10⁻¹²)/0.25
   = (0.054)/0.25
   = 0.216 N (attractive).
2. Force varies as 1/r². Doubling r makes force 1/4 of original.
3. F = (9×10⁹)(4×10⁻⁶)(1×10⁻⁶)/(0.20)²
   = (9×10⁹)(4×10⁻¹²)/0.04
   = 0.036/0.04
   = 0.90 N (repulsive, on q2 away from q1).
4. On qB: force from qA is attractive to the left:
   FBA = (9×10⁹)(2×10⁻⁶)(1×10⁻⁶)/(0.30)²
   = (9×10⁹)(2×10⁻¹²)/0.09
   = 0.018/0.09
   = 0.20 N left.
   Force from qC is attractive to the right:
   FBC = (9×10⁹)(3×10⁻⁶)(1×10⁻⁶)/(0.30)²
   = 0.027/0.09
   = 0.30 N right.
   Net = 0.10 N to the right.
5. For identical charges q: F = k q²/r² → q = √(Fr²/k)
   = √(0.90×0.10² / 9×10⁹)
   = √(0.009 / 9×10⁹)
   = √(1×10⁻¹²)
   = 1.0×10⁻⁶ C = 1.0 μC.

Set 2 Answers

1. E = kq/r² = (9×10⁹)(5×10⁻⁶)/(0.40)²
   = (45×10³)/0.16
   = 2.81×10⁵ N/C.
2. F = qE = (2.0×10⁻⁹)(3.0×10⁴) = 6.0×10⁻⁵ N.
3. V = kq/r = (9×10⁹)(−6×10⁻⁶)/0.25
   = (−54×10³)/0.25
   = −2.16×10⁵ V.
4. W = qΔV = q(Vf − Vi) = (3.0×10⁻⁹)(45 − 120)
   = (3.0×10⁻⁹)(−75)
   = −2.25×10⁻⁷ J.
   (Negative means the field does the work; required external work magnitude is 2.25×10⁻⁷ J.)
5. Between like charges, E cancels where k(2)/x² = k(8)/(0.30 − x)².
   Simplify: (0.30 − x)² = 4x² → 0.30 − x = 2x → x = 0.10 m from the +2.0 μC charge.

Set 3 Answers

1. C = Q/V = (6.0×10⁻⁶)/12 = 5.0×10⁻⁷ F = 0.50 μF.
2. C = ε₀A/d = (8.85×10⁻¹²)(0.020)/(1.0×10⁻³)
   = 1.77×10⁻¹⁰ F = 177 pF.
3. U = 1/2 CV² = 0.5(4.0×10⁻⁶)(9.0)²
   = 2.0×10⁻⁶×81
   = 1.62×10⁻⁴ J.
4. 1/Ceq = 1/3.0 + 1/6.0 = 1/2 → Ceq = 2.0 μF.
5. Ceq = 2.0 + 5.0 = 7.0 μF, so Q = CV = (7.0×10⁻⁶)(10) = 7.0×10⁻⁵ C = 70 μC.

Set 4 Answers

1. I = V/R = 24/6.0 = 4.0 A.
2. V = IR = (2.5)(3.0) = 7.5 V.
3. R = V/I = 120/0.80 = 150 Ω.
4. Rtotal = 4.0 + 8.0 = 12 Ω, so I = 12/12 = 1.0 A.
   Voltage drops: V4 = IR = (1.0)(4.0) = 4.0 V; V8 = (1.0)(8.0) = 8.0 V.
5. 1/Req = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6 = 1/2 → Req = 2.0 Ω.
   Itotal = V/Req = 12/2.0 = 6.0 A.

Set 5 Answers

1. P = I²R = (1.5)²(10) = 2.25×10 = 22.5 W.
2. I = P/V = 60/120 = 0.50 A.
3. R = V²/P = 220²/1500 = 48400/1500 ≈ 32.3 Ω.
4. Energy = Power × time = 0.100 kW × 6.0 h = 0.60 kWh.
5. Parallel part: 1/Rp = 1/3.0 + 1/6.0 = 1/3 + 1/6 = 1/2 → Rp = 2.0 Ω.
   Total R = 2.0 (series) + 2.0 (parallel eq) = 4.0 Ω.
   Itotal = 9/4.0 = 2.25 A.

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