PhysicsElectrostaticsClass 12

Electric field lines: interactive simulation

Field lines start on positive charges and end on negative ones. Where they bunch together the field is strong; where they spread out it is weak. They never cross, because the field has only one direction at each point.

Useful for: JEE Main, NEET, AP Physics 2

Try this

Each line here is traced by following the total field — the vector sum of every charge’s Coulomb field — in small steps. Tap the stage to drop a probe and read |E| and V at that point.

  1. 1

    Start from Dipole and look between the charges.

    What you should see

    Lines leave + and curve into −. The field midway points from + towards −.

  2. 2

    Choose Like charges and probe the exact midpoint.

    What you should see

    |E| is almost zero there — a neutral point — and no line passes through it.

  3. 3

    Raise one charge from 2 μC to 4 μC.

    What you should see

    Twice as many lines start on it: line count is proportional to charge.

  4. 4

    With a single charge, probe at one distance and then at double the distance.

    What you should see

    |E| drops to a quarter (inverse square), while V only halves.

Formulas behind it

Field of a point charge

E=kqr2,k=9×109 N m2C−2E = \frac{kq}{r^2}, \quad k = 9\times10^{9}\ \text{N m}^2\text{C}^{-2}

Superposition

E⃗=E⃗1+E⃗2+…\vec E = \vec E_1 + \vec E_2 + \dots

Potential of a point charge

V=kqr,Vtotal=∑kqiriV = \frac{kq}{r}, \quad V_{\text{total}} = \sum \frac{kq_i}{r_i}

Common mistakes

  • Adding field magnitudes instead of vectors.
  • Thinking a released charge always moves along a field line. That is true only for straight lines; on curved lines inertia carries it off.
  • Confusing where E = 0 with where V = 0 — between equal opposite charges V = 0 at the midpoint, but E is not.
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FAQ

Why do field lines never cross?

At a crossing the field would point in two directions at once, which is impossible — the total field at a point has a single direction.

Where is the neutral point for charges q and 4q of the same sign separated by d?

Setting kq/x² = k(4q)/(d − x)² gives x = d/3, measured from the smaller charge.

Can the electric field be zero where the potential is not?

Yes. Midway between two equal like charges E = 0 but V = 2kq/r, which is not zero.

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