Electric Field
Remove the charge q from near Q – is there “nothing” left at that point? Physicists say no: Q fills the space around it with an electric field, and the field pushes on any charge that arrives. On this page you probe the field with a test charge, map the field of several charges, race an electron against a proton, and solve Examples 1.8 and 1.9 in exam layout.
- Coulomb's law F = 14πε0Qqr2, with 1/4πε0 ≈ 9 × 109 N m2 C−2; forces from several charges add as vectors (superposition, page 8).
- Newton's second law a = F/m and, from rest under a constant acceleration, h = ½ a t2, so t = √(2h/a) (page 3).
- 1 cm = 10−2 m, 1 nC = 10−9 C, 1 μC = 10−6 C.
1. What is an electric field? (Fig 1.11)
Place a point charge Q at the origin O and another point charge q at P, with OP = r. Q pushes or pulls q according to Coulomb's law. Now take q away. Is there nothing at P? If there were nothing, how could a force suddenly act the moment we put q back? To answer this, scientists introduced the idea of a field: the charge Q produces an electric field (വൈദ്യുത മണ്ഡലം) everywhere around it. When a charge q is brought to P, the field at P acts on it and produces the force.
F(r) = q E(r) (1.8) ⇒ E = Fq unit: N/C (N C−1)
The electric field at a point is the force experienced by a unit positive charge placed at that point (without disturbing the charges that produce the field). It is a vector; its SI unit is N/C (later you will meet the equivalent unit V/m).
E = limq → 0 Fq (1.9)
The charge Q that produces the field is the source charge; the charge q that “tests” it is the test charge (പരീക്ഷണ ചാർജ് · സ്രോതസ്സ് ചാർജ്). Why the limit q → 0? A real test charge would push Q and make it move, changing the very field we want to measure. Making q negligibly small makes F negligibly small too, but the ratio F/q stays finite – that ratio is E. (In practice the source charges are simply held in place by other forces, like the charges fixed on a charged sheet.)
2. Field of a point charge
E(r) = 14πε0 Qr2 r̂ (1.6) where r̂ = r/r is the unit vector (യൂണിറ്റ് സദിശം) from Q to the point
Multiply by q and you get back Coulomb's law, F = 14πε0Qqr2r̂ (1.7). Important remarks from the book:
- If q is unity, the field is numerically equal to the force – hence “force on a unit positive charge”.
- E does not depend on q. F is proportional to q, so F/q is independent of q. E depends only on the source charge and the position r; the field exists at every point in space, whether or not a test charge is there.
- For a positive source charge E points radially outwards; for a negative one it points radially inwards.
- E depends only on the distance r, so it has the same magnitude everywhere on a sphere centred on the charge – the field of a point charge has spherical symmetry.
Drag the test charge. Change q: the force arrow (purple) grows, shrinks or flips, but the field arrow (green) does not change – E belongs to the point, not to the test charge. Make q negative: the force turns opposite to E. 1 grid square = 10 cm.
3. Field due to a system of charges (Fig 1.12)
For charges q1, q2, …, qn, the field at a point P is again the force on a unit positive test charge at P. By the superposition principle (അധ്യാരോപണ തത്വം), each charge produces its own field as if the others were absent, and the fields add as vectors (സദിശം):
E(r) = E1 + E2 + … + En = 14πε0 ∑i = 1n qiriP2 r̂iP (1.10)
Here riP is the distance from qi to P and r̂iP the unit vector from qi towards P. E is a vector that varies from point to point and is fixed by the positions of the source charges.
Drag the charges and the yellow probe ⊕. A charge set to 0 is switched off. Grid squares are 10 cm. “Log scale” makes arrows grow with log E so that weak-field regions are still visible; “∝ E (capped)” draws lengths proportional to E, as in Fig 1.15 – the arrows then vanish quickly with distance (1/r²).
4. Physical significance of the electric field
For electrostatics (charges at rest) the field is convenient but not strictly necessary – you could always compute forces directly with Coulomb's law and superposition. The field is an elegant way to describe the electrical environment of a system of charges: at each point it tells you the force a unit positive test charge would feel there. It is a property of the source charges alone, defined at every point, and – since force is a vector – it is a vector field.
When charges accelerate, no signal can travel faster than light (c = 3 × 108 m/s). So if q1 is suddenly moved, q2 far away cannot feel the change instantly. The field picture explains this delay: the accelerated q1 sends out electromagnetic waves that travel at speed c, reach q2 and then exert the force. Fields therefore have their own dynamics and can carry energy (and momentum) – a source switched on briefly leaves behind fields still travelling outward. The field idea was introduced by Faraday and is now among the central ideas of physics.
Sunlight reaching Earth after about 8 minutes, a mobile phone talking to a tower, a radio picking up a station – in each case energy is carried across space by electric and magnetic fields, with nothing material travelling between source and receiver. Closer to this chapter: a photocopier or laser printer uses the electric field of a charged drum to pull toner powder onto exactly the right places.
5. Example 1.8 – an electron and a proton falling in a field (Fig 1.13)
Example 1.8. An electron falls through a distance of 1.5 cm in a uniform electric field (ഏകീകൃത വൈദ്യുത മണ്ഡലം) of magnitude 2.0 × 104 N C−1 [Fig 1.13(a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig 1.13(b)]. Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.
Force = eE (gravity is negligible – check it at the end). Acceleration a = eE/m. Starting from rest, h = ½ a t², so t = √(2h/a) = √(2hm/(eE)). Use e = 1.6 × 10−19 C, me = 9.11 × 10−31 kg, mp = 1.67 × 10−27 kg and h in metres.
Slowed down about 30 million times: the proton's 125 ns fall takes 4 s on screen. The electron, about 1836 times lighter, crosses in only 2.9 ns – it seems to arrive instantly.
6. Example 1.9 – field at three points near two charges (Fig 1.14)
Example 1.9. Two point charges q1 and q2, of magnitude +10−8 C and −10−8 C respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig 1.14. (A is midway between the charges; B is 0.05 m to the left of q1 on the line; C is 0.1 m from each charge.)
Field of each charge: E = 9 × 109 × 10−8/r2. Direction: away from + (q1), towards − (q2). At A both point the same way; at B they oppose; at C the vertical parts cancel and the horizontal parts (E cos 60°) add.
- 🧪 Charges and Fields (PhET simulation) PhET, University of Colorado Boulder
Place charges and see the electric field vectors and field of a dipole. - 🧪 Electric Field Hockey (PhET simulation) PhET, University of Colorado Boulder
A game: steer a charged puck using the forces of other charges. - 📄 Electric Field Lines The Physics Classroom
Rules for drawing field lines, with diagrams. - 📄 Electric field Wikipedia
Definition, field of a point charge and superposition.
7. Check yourself
- Electric field = force per unit positive test charge: E = limq→0 F/q; F = qE; unit N/C. E does not depend on the test charge.
- Point charge: E = (1/4πε0)(Q/r²) r̂ – outward for +Q, inward for −Q, same size on any sphere around Q.
- System of charges: E = E1 + E2 + … (vector sum, superposition).
- The field is physically real: with time-dependent charges, fields travel at speed c and carry energy.
- Ex 1.8: te = 2.9 × 10−9 s, tp = 1.3 × 10−7 s – the heavier particle is slower, unlike free fall. Ex 1.9: EA = 7.2 × 104 N/C (right), EB = 3.2 × 104 N/C (left), EC = 9 × 103 N/C (right).