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Physics: Electricity and Magnetism - Static Electricity and Electric and Magnetic Fields

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electric Charge: Matter is made of atoms containing protons (positive charge, +e+e) and electrons (negative charge, −e-e). The SI unit for charge (QQ) is the Coulomb (CC).

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Law of Electrostatics: Like charges repel each other, while opposite charges attract each other (+↔++ \leftrightarrow + and +→←−+ \rightarrow \leftarrow -).

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Charging by Friction: When two different insulators are rubbed together, electrons are transferred from one to the other. The object that loses electrons becomes positively charged, and the one that gains electrons becomes negatively charged (Q=n×eQ = n \times e).

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Electric Fields: An electric field is a region around a charged particle where a force would be exerted on other charges. Field lines always point away from positive charges (++) and towards negative charges (−-).

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Magnetic Poles: Every magnet has two poles: North (NN) and South (SS). Like poles repel, and opposite poles attract.

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Magnetic Fields: The region around a magnet where magnetic forces can be detected. Field lines emerge from the North pole and enter the South pole.

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Electromagnetism: An electric current (II) flowing through a conductor creates a magnetic field. The strength of an electromagnet can be increased by increasing the current (II), increasing the number of turns in the coil (NN), or adding a ferromagnetic core (like iron).

📐Formulae

Q=n×eQ = n \times e

F∝q1×q2r2F \propto \frac{q_1 \times q_2}{r^2}

B∝I×NB \propto I \times N

💡Examples

Problem 1:

A plastic rod is rubbed with a cloth and loses 4×10134 \times 10^{13} electrons. If the charge of one electron is e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}, calculate the total charge acquired by the rod.

Solution:

Using the formula Q=n×eQ = n \times e: Q=(4×1013)×(1.6×10−19 C)Q = (4 \times 10^{13}) \times (1.6 \times 10^{-19} \text{ C}) Q=6.4×10−6 CQ = 6.4 \times 10^{-6} \text{ C}

Explanation:

The rod loses electrons, which are negative. Therefore, the rod becomes positively charged. The total charge is the product of the number of electrons transferred and the elementary charge.

Problem 2:

A static experiment measures an initial charge of 850 units850 \text{ units} on a sphere. After grounding, the charge drops to 275 units275 \text{ units}. Calculate the amount of charge that flowed to the ground using vertical subtraction.

Solution:

850−275575\begin{array}{r} 850 \\ - 275 \\ \hline 575 \end{array}

Explanation:

To find the charge transferred, subtract the final charge from the initial charge. The result is 575 units575 \text{ units}.

Problem 3:

How does the magnetic field strength change if the number of turns in a solenoid is tripled while the current remains constant?

Solution:

Bnew=3×BoriginalB_{new} = 3 \times B_{original}

Explanation:

Magnetic field strength (BB) in an electromagnet is directly proportional to the number of turns (NN). Tripling NN triples the strength of the magnetic field.