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Electrostatics - Electric Charges and Fields

Grade 12ICSEPhysics

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

๐Ÿ”‘Concepts

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Quantization of Charge: Charge qq on a body is always an integral multiple of the elementary charge ee, expressed as q=neq = ne, where e=1.6ร—10โˆ’19Ce = 1.6 \times 10^{-19} C.

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Coulomb's Law: The electrostatic force FF between two point charges q1q_1 and q2q_2 separated by a distance rr in vacuum is F=14ฯ€ฯต0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}, where ฯต0\epsilon_0 is the permittivity of free space (8.854ร—10โˆ’12C2Nโˆ’1mโˆ’28.854 \times 10^{-12} C^2 N^{-1} m^{-2}).

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Electric Field Intensity: The electric field Eโƒ—\vec{E} at a point is the force experienced per unit positive test charge q0q_0, i.e., Eโƒ—=Fโƒ—q0\vec{E} = \frac{\vec{F}}{q_0}. For a point charge, E=14ฯ€ฯต0qr2E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}.

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Electric Dipole: A pair of equal and opposite charges qq and โˆ’q-q separated by a small distance 2a2a. The dipole moment is pโƒ—=qร—(2aโƒ—)\vec{p} = q \times (2\vec{a}), directed from negative to positive charge.

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Electric Flux: The total number of electric field lines crossing a given area AA. It is given by ฮฆE=Eโƒ—โ‹…Aโƒ—=EAcosโกฮธ\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta.

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Gauss's Law: The total electric flux ฮฆE\Phi_E through any closed surface (Gaussian surface) is equal to 1ฯต0\frac{1}{\epsilon_0} times the net charge qq enclosed by the surface: โˆฎEโƒ—โ‹…dAโƒ—=qenclosedฯต0\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}.

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Field due to an Infinitely Long Straight Wire: The electric field at a distance rr from a wire with linear charge density ฮป\lambda is E=ฮป2ฯ€ฯต0rE = \frac{\lambda}{2\pi\epsilon_0 r}.

๐Ÿ“Formulae

q=neq = ne

F=14ฯ€ฯต0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}

Eโƒ—=Fโƒ—q0\vec{E} = \frac{\vec{F}}{q_0}

Eaxial=14ฯ€ฯต02pr(r2โˆ’a2)2โ‰ˆ14ฯ€ฯต02pr3E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}

Eequatorial=14ฯ€ฯต0p(r2+a2)3/2โ‰ˆ14ฯ€ฯต0pr3E_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}

ฯ„โƒ—=pโƒ—ร—Eโƒ—=pEsinโกฮธ\vec{\tau} = \vec{p} \times \vec{E} = pE \sin \theta

ฮฆE=โˆฎEโƒ—โ‹…dAโƒ—=qinฯต0\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\epsilon_0}

Esheet=ฯƒ2ฯต0E_{sheet} = \frac{\sigma}{2\epsilon_0}

๐Ÿ’กExamples

Problem 1:

Calculate the electrostatic force between two protons separated by a distance of 1.6ร—10โˆ’15m1.6 \times 10^{-15} m. (Charge of proton e=1.6ร—10โˆ’19Ce = 1.6 \times 10^{-19} C)

Solution:

Using Coulomb's Law: F=14ฯ€ฯต0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}. Given q1=q2=1.6ร—10โˆ’19Cq_1 = q_2 = 1.6 \times 10^{-19} C and r=1.6ร—10โˆ’15mr = 1.6 \times 10^{-15} m. F=(9ร—109)(1.6ร—10โˆ’19)2(1.6ร—10โˆ’15)2=9ร—109ร—2.56ร—10โˆ’382.56ร—10โˆ’30=90NF = (9 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{(1.6 \times 10^{-15})^2} = 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{2.56 \times 10^{-30}} = 90 N.

Explanation:

The force is calculated by substituting the charge of the protons and the distance into the Coulombic equation. The large value (90N90 N) indicates the strength of the repulsion at nuclear distances.

Problem 2:

An electric dipole with dipole moment 4ร—10โˆ’9Cโ‹…m4 \times 10^{-9} C \cdot m is aligned at 30โˆ˜30^\circ with the direction of a uniform electric field of magnitude 5ร—104N/C5 \times 10^4 N/C. Calculate the magnitude of the torque acting on the dipole.

Solution:

Torque ฯ„=pEsinโกฮธ\tau = pE \sin \theta. Given p=4ร—10โˆ’9Cโ‹…mp = 4 \times 10^{-9} C \cdot m, E=5ร—104N/CE = 5 \times 10^4 N/C, and ฮธ=30โˆ˜\theta = 30^\circ. ฯ„=(4ร—10โˆ’9)ร—(5ร—104)ร—sinโก(30โˆ˜)=20ร—10โˆ’5ร—0.5=10โˆ’4Nโ‹…m\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin(30^\circ) = 20 \times 10^{-5} \times 0.5 = 10^{-4} N \cdot m.

Explanation:

Torque is a vector product of the dipole moment and the electric field. Since sinโก(30โˆ˜)=0.5\sin(30^\circ) = 0.5, the resulting torque is 10โˆ’4Nโ‹…m10^{-4} N \cdot m.

Electric Charges and Fields - Revision Notes & Key Formulas | ICSE Class 12 Physics