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Electric Charges and Fields - Electric Field and Field Lines

Grade 12CBSEPhysics

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

🔑Concepts

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Electric Field: It is the environment around a charge where another charge experiences an electrostatic force. The Electric Field Intensity E⃗\vec{E} is defined as the force experienced per unit positive test charge q0q_0, given by E⃗=F⃗q0\vec{E} = \frac{\vec{F}}{q_0}.

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Electric Field due to a Point Charge: For a source charge QQ, the magnitude of the field at a distance rr is inversely proportional to the square of the distance, following the Inverse Square Law.

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Principle of Superposition: The net electric field at a point due to a system of point charges is the vector sum of the electric fields produced by each charge individually: E⃗net=E⃗1+E⃗2+...+E⃗n\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + ... + \vec{E}_n.

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Electric Field Lines: These are continuous curves used to represent the electric field visually. The tangent to a field line at any point gives the direction of E⃗\vec{E} at that point. They never intersect because at the point of intersection, there would be two directions of the electric field, which is impossible.

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Properties of Field Lines: They start from positive charges and end at negative charges. The density of field lines in a region indicates the relative strength of the electric field.

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Electric Dipole: A pair of equal and opposite charges qq and −q-q separated by a distance 2a2a. The electric dipole moment p⃗\vec{p} is a vector directed from −q-q to +q+q.

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Electric Field of a Dipole: The field at an axial point (distance rr) is directed along the dipole moment, while at an equatorial point, it is opposite to the dipole moment. For a short dipole (r≫ar \gg a), the axial field is twice the equatorial field.

📐Formulae

E⃗=lim⁡q0→0F⃗q0\vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0}

E=14πϵ0qr2E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}

p⃗=q×(2a⃗)\vec{p} = q \times (2\vec{a})

Eaxial=14πϵ02pr(r2−a2)2≈14πϵ02pr3 (for r≫a)E_{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} \text{ (for } r \gg a)

Eequatorial=14πϵ0p(r2+a2)3/2≈14πϵ0pr3 (for r≫a)E_{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} \text{ (for } r \gg a)

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

U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p} \cdot \vec{E} = -pE \cos \theta

💡Examples

Problem 1:

Calculate the magnitude of the electric field at a point 30 cm30\text{ cm} from a point charge of 5×10−9 C5 \times 10^{-9}\text{ C} in vacuum.

Solution:

Given: q=5×10−9 Cq = 5 \times 10^{-9}\text{ C}, r=30 cm=0.3 mr = 30\text{ cm} = 0.3\text{ m}, and 14πϵ0=9×109 N⋅m2C−2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}. Using E=14πϵ0qr2E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}: E=(9×109)5×10−9(0.3)2=450.09=500 N/CE = (9 \times 10^9) \frac{5 \times 10^{-9}}{(0.3)^2} = \frac{45}{0.09} = 500\text{ N/C}.

Explanation:

The electric field is calculated using the point charge formula. Ensure distance is converted to meters (mm) to maintain SI units.

Problem 2:

An electric dipole with dipole moment 4×10−9 C⋅m4 \times 10^{-9}\text{ C}\cdot\text{m} is aligned at 30∘30^\circ with the direction of a uniform electric field of magnitude 5×104 N/C5 \times 10^4\text{ N/C}. Calculate the magnitude of the torque acting on the dipole.

Solution:

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

Explanation:

Torque is a result of the couple acting on the two charges of the dipole in a uniform external field. It is maximum when θ=90∘\theta = 90^\circ and zero when θ=0∘\theta = 0^\circ or 180∘180^\circ.

Electric Field and Field Lines Class 12 Notes & Examples