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Electrostatics - Electric Charges and Fields

Grade 12ICSEPhysics

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

🔑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=1.6×10−19Ce = 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=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}, where ϵ0\epsilon_0 is the permittivity of free space (8.854×10−12C2N−1m−28.854 \times 10^{-12} C^2 N^{-1} m^{-2}).

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Electric Field Intensity: The electric field E⃗\vec{E} at a point is the force experienced per unit positive test charge q0q_0, i.e., E⃗=F⃗q0\vec{E} = \frac{\vec{F}}{q_0}. For a point charge, E=14πϵ0qr2E = \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 −q-q separated by a small distance 2a2a. The dipole moment is p⃗=q×(2a⃗)\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 ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta.

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Gauss's Law: The total electric flux ΦE\Phi_E through any closed surface (Gaussian surface) is equal to 1ϵ0\frac{1}{\epsilon_0} times the net charge qq enclosed by the surface: ∮E⃗⋅dA⃗=qenclosedϵ0\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 λ\lambda is E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}.

📐Formulae

q=neq = ne

F=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}

E⃗=F⃗q0\vec{E} = \frac{\vec{F}}{q_0}

Eaxial=14πϵ02pr(r2−a2)2≈14πϵ02pr3E_{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=14πϵ0p(r2+a2)3/2≈14πϵ0pr3E_{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}

τ⃗=p⃗×E⃗=pEsin⁡θ\vec{\tau} = \vec{p} \times \vec{E} = pE \sin \theta

ΦE=∮E⃗⋅dA⃗=qinϵ0\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\epsilon_0}

Esheet=σ2ϵ0E_{sheet} = \frac{\sigma}{2\epsilon_0}

💡Examples

Problem 1:

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

Solution:

Using Coulomb's Law: F=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}. Given q1=q2=1.6×10−19Cq_1 = q_2 = 1.6 \times 10^{-19} C and r=1.6×10−15mr = 1.6 \times 10^{-15} m. F=(9×109)(1.6×10−19)2(1.6×10−15)2=9×109×2.56×10−382.56×10−30=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 4×10−9C⋅m4 \times 10^{-9} C \cdot m is aligned at 30∘30^\circ with the direction of a uniform electric field of magnitude 5×104N/C5 \times 10^4 N/C. Calculate the magnitude of the torque acting on the dipole.

Solution:

Torque τ=pEsin⁡θ\tau = pE \sin \theta. Given p=4×10−9C⋅mp = 4 \times 10^{-9} C \cdot m, E=5×104N/CE = 5 \times 10^4 N/C, and θ=30∘\theta = 30^\circ. τ=(4×10−9)×(5×104)×sin⁡(30∘)=20×10−5×0.5=10−4N⋅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 sin⁡(30∘)=0.5\sin(30^\circ) = 0.5, the resulting torque is 10−4N⋅m10^{-4} N \cdot m.