Electric Field & Dipole Electric Field & Dipole E-Field vs Potential Geometry E (Electric Field) V (Potential) Remember that Field E is the negative gradient of Potential V , so V usually has one less power of r in the denominator than E . Point Charge ( q ) E = 1 4 0 q r 2 V = 1 4 0 q r Conducting Sphere (Outside, r R ) E = kQ r 2 V = kQ r Conducting Sphere (Inside, r < R ) E = 0 V = kQ R (Constant) Infinite Line Charge ( ) E = 2 0 r V = - 2 0 (r) + C Infinite Thin Sheet ( ) E = 2 0 (Uniform) V = V 0 - r 2 0 Solid Non-Conducting Sphere (Inside, r < R ) E = kQr R 3 V = kQ(3R 2 -r 2 ) 2R 3 Electric Dipole (Axial Point) E = 2kp r 3 V = kp r 2 Electric Dipole (Equatorial Point) E = kp r 3 V = 0 e field vs potential Every charge changes the space around it. Place a tiny positive test charge near a source charge and it feels a push or a pull. We call that "push per unit charge" the electric field at that point. The field exists even when no test charge is present; it is a property of space due to sources. This viewpoint is powerful: once you know the field everywhere, you can predict forces on any charge with a simple multiplication. Electric fields have direction and magnitude (they are vectors), and fields from different charges add by the head-to-tail rule. To picture the shape of a field, we use field lines: they start on positive charges and end on negative charges, are denser where the field is strong, and never cross. Many NEET problems rely on symmetry: for a point charge the field is radial and falls as 1/ r 2 ; for a long line of charge it falls as 1/r; for an infinite sheet it is constant. A particularly important configuration is an electric dipole: two equal and opposite charges separated by a small distance. At large distances a dipole’s field falls as 1/ r 3 , much faster than a single charge. In a uniform external field a dipole does not translate but tends to rotate; the torque and the potential energy depend on its orientation. Master these patterns and you can quickly reduce complex charge distributions into manageable pieces. remember Analogy: Earth’s gravity is a field. You do not need to drop a stone to know gravity exists at your location. Similarly, the electric field exists due to charges whether or not a test charge is present. Force per unit positive test charge at a point. Vector directed along the force on a +q test charge. Electric field (E) Test charge An imaginary, small positive charge used to probe a field without disturbing the source distribution. Superposition principle Net electric field is the vector sum of fields due to individual charges at that point. Field line Imaginary curve tangent to E at every point, starting on + charges and terminating on − charges; density indicates field strength. Two point charges +q and −q separated by a distance 2a (or vector d). Net charge is zero. Electric dipole Vector quantity p = q , d directed from the negative to the positive charge; unit C m . Dipole moment (p) Axial line of a dipole The line passing through both charges, collinear with p . The line perpendicular to the dipole axis through its center (also called the perpendicular bisector). Equatorial line of a dipole Torque on a dipole In a uniform field E , = p E ; magnitude = pE . Surface charge density ( ) Charge per unit area on a surface; unit C/m 2 . Linear charge density ( ) Charge per unit length on a line; unit C/m . Definition and measurement: If a test charge q 0 at a point experiences force F , the electric field there is E = q 0 0 F q 0 . In practice, we use a very small positive test charge so that it does not rearrange the sources. The direction of E is the direction of the force on a + test charge. If you know E , then the force on any charge q placed at that point is F = q , E . Field of a point charge Magnitude E= 1 4 0 |q| r 2 , directed radially away from +q and toward −q. Direction conventions: Field lines emerge from positive charges and enter negative charges. At any point the field vector is tangent to the field line. Around a single positive charge, vectors point radially outward; around a negative charge they point inward. For multiple charges, directions add vectorially; often it is easiest to resolve into perpendicular components and add. Electric field is force per unit test charge: E=F/q (vector form E = F /q ). E = F/q E = F q Choose a positive test charge by convention. Limit ensures non-perturbation of sources. Operational definition in experiments. Test charge is sufficiently small to avoid disturbing source charges. Electrostatic (charges at rest) conditions. Electrostatic force between two point charges: F = k e , |q 1 q 2| r 2 along the line joining them. F = k e |q 1 q 2| r 2 F = k e |q 1 q 2| r 2 From torsion-balance experiments. Try power-law with unknown n. Field lines spread over area 4 r 2 . Introduce constant k e = 1/(4 0) . Point charges, stationary. Isotropic, homogeneous space (vacuum/air). Action along the line joining charges (central force). Superposition in action: For multiple source charges q i at position vectors r i , the electric field at a point r is E ( r ) = i 1 4 0 , q i | r - r i| 2 , n i , where n i points from q i toward the field point. Practically, break each contribution into x- and y-components, add algebraically, and reconstruct the magnitude and direction. Symmetry (equal charges placed symmetrically) often cancels some components, speeding up the solution. tip Coulomb’s law boundaries: point charges or spherically symmetric charges (use center) and electrostatic conditions. For extended or irregular distributions at close range, integrate or use Gauss’s law when symmetry allows. Field lines: We draw a few representative lines to visualize direction and relative strength. They never intersect (there cannot be two distinct directions of E at the same point). Lines are closer where E is stronger. For an isolated conductor in electrostatic equilibrium, lines are perpendicular to its surface. Field line density is a picture aid, not a numeric measure; use formulas for exact values. Properties of electric field lines Originate on + charges, terminate on − charges (or at infinity). Never cross; tangent to a line gives direction of E. Density of lines ∝ magnitude of E (qualitative). E is perpendicular to a conducting surface in electrostatics. Lines do not enter a cavity in a conductor with no enclosed charge. They show direction of E at each point, not the dynamic trajectory (which depends on initial velocity and can curve differently). Field lines show the actual paths a test charge will follow. From discrete to continuous: If charge is spread continuously with density , , or , break it into small elements dq and integrate d E = 1 4 0 , dq r 2 , r . Symmetry can turn the integral into a simple algebraic step (e.g., rings, discs, lines, sheets). When symmetry is high, Gauss’s law is often faster than direct integration. Dipole basics: An electric dipole consists of +q and −q separated by vector d (from − to +). Its dipole moment is p =q , d . Far from the pair (distance r d ), the net charge looks like zero and the leading effect is the dipole term. The field pattern has loops from + to −, with strong variation near the charges and a characteristic 1/ r 3 decay at large distances. Direction from − to +; unit C m . Dipole moment This value quantifies the strength of the dipole based on the product of the charge magnitude and the separation distance. Axial vs equatorial: On the axis of a short dipole ( r d ), fields from +q and −q add along p . On the equatorial line they partly cancel and the direction is opposite to p . Remember the magnitude on the axial line is twice that on the equatorial line at the same r (short dipole approximation). Directions: axial along p (for point outside between charges), equatorial opposite to p . Dipole field (short dipole) This calculation determines the electric field strength at any point relative to the dipole, accounting for axial and equatorial positions. Axial point on x-axis; equatorial on y-axis. Along the axis; directions add. Account for opposite signs. Keep first-order in a/r. With p=q(2a) . Components perpendicular add to give the 1× factor; direction opposite p . E axial = 1 4 0 2p r 3 , E equatorial = 1 4 0 p r 3 E axial = 1 4 0 2p r 3 , ; E equatorial = 1 4 0 p r 3 Short dipole: r d (or r a with separation 2a ). Vacuum/air medium. Dipole in a uniform field: A dipole experiences no net force in a perfectly uniform field (forces on +q and −q cancel) but a torque tends to align p with E . The torque magnitude is = pE (zero at =0 ,180 ; maximum at 90 ). Its potential energy is U=- p E =-pE , minimum when aligned ( =0 ) and maximum when anti-aligned. Torque and energy of a dipole In a uniform field, net force is zero but torque and energy depend on orientation. This energy determines the stability and equilibrium positions of a dipole relative to an external uniform electric field. Stability and non-uniformity: =0 is a stable equilibrium (small rotations increase U), while 180 is unstable. In a non-uniform field, a dipole feels a net force toward the region of stronger field (because the two charges experience unequal magnitudes of force). This is the basis of dielectrophoresis and polar molecule alignment. Two significant figures for NEET-style answers. Negative indicates attraction toward +q, opposite to E. E, then F on q t Find the electric field at 20 cm from a +3.0 µC point charge in air. Also find the force on a −1.0 nC charge placed there. q = +3.0 , C = 3.0 10 -6 , C r = 0.20 , m k e = 9.0 10 9 , N ,m 2/C 2 test charge q t = -1.0 , nC = -1.0 10 -9 , C E in N/C, F in N easy Use E= k e |q| r 2 and F =q t , E with direction away from +q. q 1 = +4.0 , C , ; x 1=-0.30 , m q 2 = -2.0 , C , ; x 2=+0.30 , m k e = 9.0 10 9 , N ,m 2/C 2 N/C Two charges are on the x-axis: +4.0 µC at x = −0.30 m and −2.0 µC at x = +0.30 m. Find the net electric field at the origin. Direction: away from +q1, so from x=-0.30 m toward negative x (left). Direction: toward −q2, i.e., from origin toward +x (right). Negative means net field points to −x (left). Net E at x=0 (magnitude and direction) medium Compute E 1 at O due to q1 and E 2 due to q2. Use signs for direction along x. Two significant figures. Initial energy. Final energy (aligned). Negative means the field does positive work; external agent would need +2.5× 10 -6 J to hold/rotate oppositely. Torque τ and work W to rotate from 60° to 0° An electric dipole has q = 2.0 nC and separation d = 1.0 cm. It is placed in a uniform field E = 2.5× 10 5 N/C making 60° with the dipole axis. Find (a) torque magnitude and (b) work required to rotate it to align with the field. q = 2.0 10 -9 , C d = 1.0 10 -2 , m p = qd = 2.0 10 -11 , C ,m E = 2.5 10 5 , N/C = 60 N·m, J hard Use = pE and W= U = U final -U initial =-pE f + pE i . Electric field near a long line charge: E= 2 0 r , radial. Infinite straight wire with uniform linear charge density . Gaussian cylindrical symmetry; electrostatics. By symmetry, E is radial and constant on curved surface. No flux through end caps (E is parallel to ends). Charge enclosed by the cylinder. Apply Gauss’s law. Cancel L and solve for E. E = 2 0 r E = 2 0 r Interpreting the line-charge field: The 1/r dependence reflects cylindrical spreading of field lines; doubling the distance halves E. The direction is perpendicular to the wire, pointing outward for positive and inward for negative . Use this formula when the wire length is much larger than your distance from it (end effects negligible). Electric field of a uniformly charged infinite non-conducting sheet: E= 2 0 , independent of distance. E = 2 0 E = 2 0 Field is perpendicular to the sheet. Equal flux through both end caps; zero through curved side. Enclosed charge within the pillbox. Gauss’s law =q enc / 0 . Distance from the sheet does not appear. Infinite plane sheet with uniform (or point very close to a large sheet). Electrostatic conditions; non-conducting sheet. Why distance does not matter for a sheet: Doubling your distance changes the area sampling the field in exactly the way that preserves flux through a Gaussian pillbox. Physically, translational symmetry parallel to the plane means there is no preferred distance scale to make E vary with r. For an ideal conductor plane, just outside the surface E= / 0 (all field lines lie on one side). Field of a uniformly charged thin spherical shell: E=0 inside; E= 1 4 0 Q r 2 outside. Hence E=0 inside. Entire shell is enclosed. Gauss’s law on a spherical surface. Same as a point charge at the center. E(r) = 0 ; (r<R); E(r) = 1 4 0 Q r 2 ; (r R) E(r) = 0 ; (r<R); ; E(r)= 1 4 0 Q r 2 ; (r R) Thin spherical shell, uniform charge Q. Electrostatic conditions; spherical symmetry. Inside a charged shell, the electric field is zero everywhere, but the potential is constant (non-zero in general). A solid non-conducting sphere with uniform volume charge is different: inside, E increases linearly with r. For metals (conductors) in electrostatics, all excess charge resides on the outer surface, the interior field is zero, and the surface is an equipotential. Electric flux through a flat area: E = EA , where is angle between E and the area normal. Dot product definition. Pull constants out of the integral. E = E A E = E A Uniform E over a flat surface area A. Angle defined between E and outward normal. Flux intuition: Count field lines piercing a surface; tilt the surface and fewer lines pass through (cosθ factor). Rotate to be parallel to E and the flux becomes zero. Flux changes sign if we reverse the choice of area normal. In Gauss’s law, the net flux through a closed surface equals q enc / 0 . custom r (distance from center) E (field magnitude) N/C 0 to E0 0 to R Inside shell: E = 0 At surface: maximum Q/(4πϵ0 R 2 ) 2R Outside: inverse-square Q/(16πϵ0 R 2 ) Electric field of a uniformly charged thin spherical shell: zero inside, jump to finite value at r=R, then 1/ r 2 falloff. E vs r for a uniformly charged thin spherical shell. control control Comparison of common symmetric sources Point charge q Spherical ∝ 1/ r 2 Radial (out of +, into −) E = (1/4πϵ0)|q|/ r 2 Long line (λ) Cylindrical ∝ 1/r Perpendicular to line E = λ/(2πϵ0 r) Infinite sheet (σ) Planar Constant (independent of r) Normal to sheet (both sides) E = σ/(2ϵ0) Short dipole (p) Axial/Equatorial ∝ 1/ r 3 (far field) Axial: along p; Equatorial: opposite p E ax = (1/4πϵ0)(2p/r 3 ), E eq = (1/4πϵ0)(p/r 3 ) Source Symmetry E vs distance Direction Formula (magnitude) A dipole’s field falls like 1/ r 2 because it is made of charges. Far from the pair the net charge is zero; the leading term is dipolar and falls as 1/ r 3 . Trap: On the equatorial line of a dipole, the field direction is opposite to p . Also, a dipole in a uniform field has zero net force for any angle (only torque acts). Do not confuse “zero torque” at θ = 0° or 180° with “zero force” (force is zero regardless of orientation in a uniform field). neet-alert Axial Twice, Equatorial Half: For a short dipole at the same r, E axial = 2 ,E equatorial ; directions: axial along p , equatorial opposite p . Edge cases to watch: As r → 0 for a point charge, E (idealization). For a thin spherical shell, E has a discontinuity at the surface, consistent with surface charge density. Between large parallel plates with uniform surface charge (non-conducting sheets), the fields from each sheet add to give a nearly uniform result. Always specify the sign convention for r and for the dipole direction (− to +). tip Use the short-dipole formulas only when r d (or r 2a ). If the observation point is comparable to the separation, use exact expressions or component-wise addition of the two point-charge fields. Sketch the geometry; mark symmetry axes and likely E directions. Choose a sign convention and coordinate axes; define unit vectors. If symmetry is high (spherical/cylindrical/planar), consider Gauss’s law; else use superposition/integration. Compute components, add vectorially, and simplify. Check limits: far field (r→∞), near field, and special angles; verify units. Workflow for E-field problems Units and significant figures: The SI unit of electric field is N/C (equivalently V/m). Charge is in coulombs; microcoulomb (µC = 10 −6 C) and nanocoulomb (nC = 10 −9 C) are common. NEET numericals usually expect 2 significant figures; keep a consistent constant value (e.g., k e 9.0 10 9 , N ,m 2/C 2 ) unless otherwise specified. Recap: Electric field is the force per unit charge and adds by superposition. Field lines help visualize direction and relative strength but never replace calculation. Symmetry gives quick formulas for point, line, and sheet charges; Gauss’s law justifies them. A dipole’s field falls as 1/ r 3 and its orientation in a uniform field is governed by = p E and U=- p E . Knowing when each model applies—and its boundaries—turns tough NEET questions into quick, confident answers. Field intensity Electric field Force per unit positive test charge; vector. Vector addition of fields from multiple sources. Superposition Imaginary curve tangent to E; starts on +, ends on −. Field line Two equal and opposite charges separated by a small distance. Dipole Measure of dipole strength; p =q , d . Dipole moment Torque on dipole = p E ; tends to align p with E . Surface charge density Charge per unit area on a surface. Linear charge density Charge per unit length on a line. Φ E Measure of field lines through a surface: E=EA . Electric flux Key terms recap