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Electricity - Static Electricity

Grade 9IB

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

šŸ”‘Concepts

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Static electricity refers to the buildup of electric charge on the surface of an object. These charges remain stationary until they are able to move away by means of an electric current or electrical discharge.

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The Law of Electrostatics: Like charges repel each other (e.g., positive repels positive), and opposite charges attract each other (positive attracts negative).

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Atomic structure and charge: Protons carry a charge of +e+e, while electrons carry a charge of āˆ’e-e. Neutrons are electrically neutral. An object becomes charged by the transfer of electrons, not protons.

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Conservation of Charge: The total electric charge in an isolated system remains constant. Charge is neither created nor destroyed, only transferred from one body to another.

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Charging by Friction: When two different insulators are rubbed together, electrons are transferred from one to the other. The material that loses electrons becomes positively charged (++), and the one that gains electrons becomes negatively charged (āˆ’-).

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Charging by Induction: A method used to charge a conductor without physical contact. A charged object brought near a neutral conductor causes a redistribution of charges within the conductor.

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Electric Field: A region of space around a charged object where another charged object experiences a force. Field lines point away from positive charges and toward negative charges.

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Grounding (Earthing): The process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size, such as the Earth.

šŸ“Formulae

Q=nā‹…eQ = n \cdot e

F=kq1q2r2F = k \frac{q_1 q_2}{r^2}

E=FqE = \frac{F}{q}

šŸ’”Examples

Problem 1:

A plastic rod is rubbed with a wool cloth and acquires a net charge of āˆ’3.2Ɨ10āˆ’9Ā C-3.2 \times 10^{-9} \text{ C}. Calculate the number of electrons transferred to the rod.

Solution:

Using the formula Q=nā‹…eQ = n \cdot e, we rearrange to find nn: n=Qen = \frac{Q}{e} Given Q=āˆ’3.2Ɨ10āˆ’9Ā CQ = -3.2 \times 10^{-9} \text{ C} and the charge of one electron e=āˆ’1.6Ɨ10āˆ’19Ā Ce = -1.6 \times 10^{-19} \text{ C}, we get: n=āˆ’3.2Ɨ10āˆ’9āˆ’1.6Ɨ10āˆ’19=2.0Ɨ1010n = \frac{-3.2 \times 10^{-9}}{-1.6 \times 10^{-19}} = 2.0 \times 10^{10}

Explanation:

The negative sign indicates that electrons (which are negatively charged) were added to the rod. The result shows that 2020 billion electrons were transferred during the process.

Problem 2:

Two point charges, q1=+2Ɨ10āˆ’6Ā Cq_1 = +2 \times 10^{-6} \text{ C} and q2=+4Ɨ10āˆ’6Ā Cq_2 = +4 \times 10^{-6} \text{ C}, are separated by a distance of 0.05Ā m0.05 \text{ m}. Calculate the magnitude of the electrostatic force between them. (Use k=8.99Ɨ109Ā NĀ m2/C2k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2)

Solution:

Substitute the values into Coulomb's Law: F=kq1q2r2F = k \frac{q_1 q_2}{r^2} F=(8.99Ɨ109)(2Ɨ10āˆ’6)(4Ɨ10āˆ’6)(0.05)2F = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.05)^2} F=(8.99Ɨ109)8Ɨ10āˆ’120.0025F = (8.99 \times 10^9) \frac{8 \times 10^{-12}}{0.0025} Fā‰ˆ28.77Ā NF \approx 28.77 \text{ N}

Explanation:

Since both charges are positive, the force is repulsive. The magnitude of the force is approximately 28.77Ā N28.77 \text{ N}.