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Electric Charges and Fields - Electric Charge, Conductors and Insulators

Grade 12CBSEPhysics

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

🔑Concepts

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Electric Charge is an intrinsic property of elementary particles like electrons and protons. It is denoted by QQ or qq.

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There are two types of charges: Positive (carried by protons) and Negative (carried by electrons). Like charges repel each other, while unlike charges attract.

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Quantization of Charge: Charge on any body is an integral multiple of the basic unit of charge ee, where e=1.602×10−19 Ce = 1.602 \times 10^{-19} \text{ C}. Mathematically, Q=±neQ = \pm ne.

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Additivity of Charges: Total charge of a system is the algebraic sum of all individual charges: Q=q1+q2+⋯+qnQ = q_1 + q_2 + \dots + q_n.

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Conservation of Charge: The total charge of an isolated system remains constant. Charge can neither be created nor destroyed; it can only be transferred.

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Conductors: Materials that allow electricity to flow through them easily because they contain 'free electrons'. Metals (like Copper, Silver) and the Earth are good conductors.

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Insulators: Materials that offer high resistance to the flow of electricity because electrons are tightly bound to atoms. Examples include glass, plastic, and dry wood.

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Earthing/Grounding: The process of sharing charges with the Earth through a conductor. It is a safety measure to prevent electric shocks.

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Charging by Induction: The process of charging a neutral body by bringing a charged body near it without making physical contact.

📐Formulae

Q=±neQ = \pm ne

where n=1,2,3,… and e=1.6×10−19 C\text{where } n = 1, 2, 3, \dots \text{ and } e = 1.6 \times 10^{-19} \text{ C}

Qtotal=∑i=1nqiQ_{total} = \sum_{i=1}^{n} q_i

💡Examples

Problem 1:

A glass rod is rubbed with silk and acquires a positive charge of 4.8×10−17 C4.8 \times 10^{-17} \text{ C}. Calculate the number of electrons lost by the glass rod.

Solution:

Given Q=4.8×10−17 CQ = 4.8 \times 10^{-17} \text{ C} and e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}. Using the formula Q=neQ = ne, we have n=Qen = \frac{Q}{e}.

n=4.8×10−171.6×10−19n = \frac{4.8 \times 10^{-17}}{1.6 \times 10^{-19}} n=3×102=300n = 3 \times 10^2 = 300

Explanation:

Since the glass rod became positively charged, it must have lost electrons. By applying the quantization principle, we find that 300300 electrons were transferred from the rod to the silk.

Problem 2:

A system consists of two charges q1=+5μCq_1 = +5 \mu\text{C} and q2=−2μCq_2 = -2 \mu\text{C}. What is the total charge of the system?

Solution:

Total charge Q=q1+q2Q = q_1 + q_2. Converting to standard units: q1=5×10−6 Cq_1 = 5 \times 10^{-6} \text{ C} and q2=−2×10−6 Cq_2 = -2 \times 10^{-6} \text{ C}.

5×10−6−2×10−63×10−6\begin{array}{r} 5 \times 10^{-6} \\ - 2 \times 10^{-6} \\ \hline 3 \times 10^{-6} \end{array}

Q=3×10−6 C=3μCQ = 3 \times 10^{-6} \text{ C} = 3 \mu\text{C}

Explanation:

Electric charge is additive. We perform an algebraic sum taking the signs of the charges into account.