CLASS 12 · CHAPTER 1 · ELECTROSTATICS
Electric Charges and Fields
Rub a plastic comb through dry hair and it lifts small bits of paper without ever touching them. Something real is reaching across that gap, even though there's nothing visible in between.
Watch it happen
The little arrows show the direction of the net field everywhere in the plane, purely from adding the two charges’ fields vector by vector. Point P sits 15 cm beyond q₂, 25 cm from q₁: its field is the signed sum of both, not either charge alone.
Set both charges positive and the arrows point outward from both, away from each other everywhere nearby, repulsion. Flip one charge negative and the whole arrow pattern reorganises: arrows now curve from the positive charge into the negative one. Nothing about either charge changed individually, only the field each one produces got added to the other’s, vector by vector, at every point in the plane.
Where the formula comes from
Coulomb’s law gives the force between two point charges separated by distance :
with N·m²/C². The electric field at a point is defined as the force a tiny positive test charge would feel there, per unit charge, which makes the field due to a single point charge :
For more than one source charge, the superposition principle says the total field is simply the vector sum of each charge’s field on its own, exactly what the simulation above computes at every one of its arrows:
For an electric dipole (charges and a distance apart, dipole moment ), superposition gives a field that falls off faster than a single charge’s, once you’re far enough away ():
Where the shortcut stops working
It’s tempting to think the field only exists once you put a test charge there to feel it, as if the space were empty until then. It isn’t: the field is entirely a property of the source charges, present at every point whether or not anything is placed there to detect it. The test charge is a measuring instrument, not a precondition.
A second trap: treating Coulomb’s law as roughly inverse with distance. It is inverse-square. Pull the two charges in the simulation twice as far apart and the force drops to a quarter, not a half, drag them three times as far and it drops to a ninth.
The sharpest trap sits inside a uniformly charged spherical shell: it’s easy to assume the field there is just “very weak”, since you’re surrounded by charge on all sides. Gauss’s law says it is exactly zero, not approximately, because any sphere drawn inside the shell encloses no charge whatsoever. This isn’t a side detail, it’s one of the most precise experimental confirmations that the exponent in Coulomb’s law really is 2.
Apply it under exam conditions
Q1. Two charges q₁ = +2 µC and q₂ = −3 µC are 10 cm apart. Find the force between them.
Check it directly: set the simulation above to q₁ = 2, q₂ = −3, the force readout matches.
Q2. A thin spherical shell of radius 10 cm carries a total charge of 2 µC spread uniformly over its surface. Find the field (a) 5 cm from the centre, (b) 20 cm from the centre.
Quick answers
If I take away the test charge, does the electric field disappear too?+
No. The field is a property of the source charges alone, it exists at every point in space whether or not anything is there to feel it. A test charge is only how we detect and measure it; removing it changes nothing about the field itself, the same way a road doesn't vanish when no car is driving on it.
Does doubling the distance between two charges halve the force between them?+
No, it quarters it. Coulomb's law is an inverse-square law, F is proportional to 1/r2, not 1/r. Doubling r multiplies the force by 1/4, tripling it multiplies the force by 1/9.
Is the electric field inside a uniformly charged spherical shell really exactly zero, or just very small?+
Exactly zero, everywhere inside, not an approximation. It follows directly from Gauss's law: any spherical Gaussian surface drawn inside the shell encloses no charge at all, so the enclosed-charge side of Gauss's law is identically zero. This result is sensitive enough that experimentally confirming it is one of the best tests that the exponent in Coulomb's law is precisely 2, not 2.00001 or 1.99999.
Why does a dipole's field fall off faster (as 1/r3) than a single charge's field (as 1/r2)?+
Because at large distances the two opposite charges' fields nearly cancel, only a small residual survives, and that residual shrinks faster with distance than either charge's own field does. A single charge has nothing to cancel against, so it keeps its full 1/r2 fall-off.
Is electric flux the same thing as the electric field?+
No. The electric field E is a vector defined at every point in space. Electric flux is a single number for a given surface, roughly 'how much field pierces through that surface', computed from E, the surface's area, and the angle between them. Field is local and point-by-point; flux is a property of a whole surface.
Related concepts
Physics doesn’t stay inside chapter boundaries. Neither should you.