Derivations and worked examples: flux, dipole and torque

Electric Charges and Fields24 min readLesson 4 of 8

4.1 Electric flux Think of flux as the amount of field that pierces a surface. Tilt the surface and less field passes through it. Measure the tilt by the angle between the field and the outward normal to the surface. The flux is the dot product of the field and the area vector, and its unit is N m^2 C^-1. For a closed surface the outward normal is positive everywhere. Field leaving the surface counts as positive flux, and field entering counts as negative flux. A charge placed outside a closed surface sends in as much field as it sends out, so its net contribution to the flux is zero. Only charges inside the surface change the net flux. 4.2 The electric dipole Two equal charges of opposite sign separated by a small distance 2a form an electric dipole. The dipole moment has size p = q \times 2a and points from the negative charge toward the positive charge. The moment is a vector: its direction follows the axis of the pair, not the field. The field of a dipole is worth deriving because it appears in later chapters too. Take a point on the axis at distance r from the centre, with r much larger than a. The field of the positive charge is k q / (r - a)^2 away from it, and the field of the negative charge is k q / (r + a)^2 toward it. Subtract, use r much larger than a, and keep the first non-zero term. The two square terms nearly cancel and what remains goes as 2k p / r^3. On the equatorial plane the two fields are equal in size and the angle between them adds a cosine factor. The result is k p / r^3, half the axial field for the same distance, and the net field points opposite to the dipole moment.

ϕ = E \cdot S = E S cos θ

Flux through a planar surface; theta is the angle between E and the area vector

\oint E \cdot d S = \frac{q _{enc}}{ε _{0}}

Flux through a closed surface depends only on the enclosed charge

p = q (2 a)

Dipole moment, directed from the negative to the positive charge, unit C m

E_{axial} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}, E_{equatorial} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}} (r ≫ a)

Far field of a dipole on the axis and on the equatorial plane

Two diagrams of the electric field of a dipole, first with the point on the axis and then with the point on the equatorial plane, showing the field vectors and the dipole moment.

4.3 Torque and potential energy Place a dipole in a uniform field. The forces on the two charges are equal in size and opposite in direction, so the net force is zero and the centre of the dipole does not move. The pair of forces does form a couple, and the couple turns the dipole. The lever arm is a sin theta on each side, so the torque has size p E sin theta. In vector form the torque is the cross product of the dipole moment and the field. The torque is largest when the dipole is perpendicular to the field and zero when it is parallel or antiparallel. Work done by the field in turning the dipole is stored as potential energy, and the standard expression U = -p E cos theta is lowest when the moment points along the field. A dipole aligned with the field is stable, and an antiparallel dipole is unstable. 4.4 Worked example: torque and energy A dipole of moment 4 \times 10^-9 C m is held at 30 degrees to a uniform field of 5 \times 10^4 N C^-1. The torque is p E sin 30, which is 4 \times 10^-9 \times 5 \times 10^4 \times 0.5, so 1 \times 10^-4 N m. The potential energy is -p E cos 30, which is about -1.73 \times 10^-4 J. A second standard problem gives two charges, one at (0,0) and one at (3 mm, 4 mm), and asks for the torque in a field of 1000 N C^-1 along the x-axis. Find the separation from the coordinates, multiply the charge by that separation to get p, then use the cross product with the field. Keep the direction in the answer: the torque vector is perpendicular to both p and E. 4.5 The continuous limit A very small element of a charged body is still a dipole when a second charge sits nearby, because the field of that second charge varies across the element. The force on the element is then the force on its net charge plus the force on its induced dipole. This picture becomes the polarisation of a dielectric, which returns in the capacitance chapter.

A dipole placed at an angle in a uniform electric field, with the equal and opposite forces on its two charges forming a couple that rotates the dipole.

Exam

Write the far field of a dipole only with the phrase r much greater than a. For distances comparable to the dipole size the derivation is not valid and the two denominator terms must be kept. A numerical answer without the direction of the torque or field loses marks.

4.6 Common mistakes in this topic Three errors appear again and again. Writing E = k p / r^2 instead of k p / r^3 forgets that the two charges nearly cancel. Using the axial formula for an equatorial point, or the other way round, swaps the factor of two. And adding torque magnitudes without checking the direction leaves the answer half finished, because torque is a vector. Keep the formulas labelled axial and equatorial until they are automatic.

in hinglish

Electric flux ka matlab hai field kitna surface se guzarta hai. Formula E S cos theta hai, aur closed surface ke liye bahar ki normal positive hoti hai. Andar ka charge hi net flux decide karta hai. Dipole do barabar aur ulti charges se banta hai, aur dipole moment negative se positive ki taraf hota hai, p = q \times 2a . Axis wale point par far field 2kp / r^3 hota hai aur equatorial point par kp / r^3, isliye axial field double hota hai. Direction equatorial plane par dipole moment ke opposite hoti hai. Uniform field mein dipole par net force zero hota hai par torque p E sin theta lagta hai, isliye dipole ghoom kar field ke along aata hai. Potential energy -p E cos theta hoti hai, jo parallel position par minimum hoti hai. Numericals mein separation coordinates se nikaalo, units badlo, aur torque ki direction cross product se likho. Sirf magnitude likhna marks katwa deta hai.