Electric Field Visualiser

Place charges and watch electric field lines and equipotentials form in real time.

Starting values — Electric Field Visualiser

Charges (q,x,y; ...)	1,0,0
Field lines	on
Equipotentials	off

Change any control above and the results recompute from these.

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Some paths through this field are completely free 🖖

Switch the equipotentials on and the picture gains a second family of curves. Sliding a charge along any one of those rings costs exactly zero work — not a small amount, zero — however far round you go, because every point on the ring sits at the same potential. That is what makes the field conservative, and it has a sharper consequence: carry a charge on any closed loop at all, however contorted, and the total work comes back to zero. Notice too that the field lines and the equipotentials always meet at right angles, everywhere, on every preset. That is not decoration. The field points the direction potential falls fastest, and the direction of steepest descent is always perpendicular to the level line you are standing on.

Making an invisible push visible 🖖

This tool turns something you cannot see — the electric influence around a charge — into a picture. Each line traces the path a tiny positive test charge would be pushed: away from positive charges, toward negative ones. The single most useful habit is to read the spacing. Where lines crowd together the field is strong; where they spread apart it is weak. Line density is force made graphic.

Why you can't trap a charge in mid-air 🖖

Place two like charges and a null point appears between them where the field is exactly zero. It looks like a resting spot — but it is a saddle, not a bowl: nudge a test charge and it escapes. Earnshaw's theorem (1842) proves that no fixed arrangement of charges can hold another charge in stable static equilibrium. The same maths forbids levitating a magnet with magnets alone — you always need spin, diamagnetism, or active control.

Problems solved in full

The dipole arrangement with two charges 4 units apart 6 steps
Two charges of equal size and opposite sign sit 4 units apart. Find the field and the potential at the midpoint between them, then find how the field behaves far away. This is the Dipole arrangement: q₁ = +1 at (-2, 0), q₂ = -1 at (+2, 0), with the probe at the origin.
1. Superposition is the whole of electrostatics: each charge builds its field as though the other were not there, and you add the vectors. The vector that matters points from the charge to the probe, which is what lets the sign of q decide whether the field runs away from the charge or back into it.
  
  $E (p) = i \sum q_{i} \frac{p - x _{i}}{∣ p - x _{i} ∣ ^{3}}, p = (0, 0)$
2. The positive charge sits 2 units to the left, so its vector to the probe points 2 units right, and the cube of 2 in the denominator gives 0.250. Positive charge, outward field: this contribution points right.
  
  $q_{1} = + 1, x_{1} = (- 2, 0) : r_{1} = (2.000, 0.000), r_{1} = 2.000, E_{1} = \frac{+ 1 \cdot ( 2 , 0 )}{2 ^{3}} = (0.250, 0.000)$
3. The negative charge sits 2 units to the right, so its vector points left — and q is negative, which flips it again. Two flips, and the contribution points right as well. Pause here: the charges are on opposite sides of the probe and carry opposite signs, and those two oppositions cancel instead of compounding.
  
  $q_{2} = - 1, x_{2} = (2, 0) : r_{2} = (- 2.000, 0.000), r_{2} = 2.000, E_{2} = \frac{- 1 \cdot ( - 2 , 0 )}{2 ^{3}} = (0.250, 0.000)$
4. So the two contributions add head to tail rather than fighting, and the pair delivers 0.500 where either charge alone would deliver 0.250.
  
  $E = E_{1} + E_{2} = (0.500, 0.000), ∣ E ∣ = 0.500$
5. The potential does the opposite thing with the same two charges. V is a scalar, so it carries the sign of q and nothing else, and the two distances are equal — the terms are +0.500 and -0.500 and the midpoint sits at V = 0. Field and potential are not two readings of one quantity: here one doubles and the other vanishes at the same point.
  
  $V = i \sum \frac{q _{i}}{r _{i}} = \frac{+ 1}{2.000} + \frac{- 1}{2.000} = 0.500 - 0.500 = 0.000$
6. Now leave the midpoint. Put the probe at height y on the perpendicular bisector, where both charges are the same distance away. The vertical components cancel by symmetry and the horizontal ones survive, giving one formula for the whole line. Set y = 0 in it and it has to reproduce the number you already have — it does, which is how you know the formula is the same physics and not a new one.
  
  $p = (0, y) : E_{y} = 0, E_{x} = \frac{2 q a}{( a ^{2} + y ^{2} ) ^{3/2}}, a = 2; y = 0 : \frac{4}{8} = 0.500$
Answer

The tool prints |E| = 0.500 at the origin, assembled from two contributions of 0.250 that point the same way, and V = 0.000 at that same point. Step 6 is the one to keep. Far out, the two 1/r² fields very nearly cancel, and what survives is set by the separation and falls as 1/y³ — one whole power of distance faster than a single charge. The ratio is 2a/y = 4/y, so at y = 20 this pair musters just under a fifth of what one lone charge of the same size would produce there. The reason is printed one row above the field: Q_total = 0.00. The 1/r² term belongs to the total charge, and when that is zero the separation is the only thing left to feel. A neutral pair is not invisible — it is short-ranged, and those are different things.

$y ≫ a : E_{x} \to \frac{2 q a}{y ^{3}} = \frac{4}{y ^{3}}, \frac{E _{x}}{q / y ^{2}} = \frac{2 a}{y} = \frac{4}{y}; y = 20 : 0.2$
Two identical positive charges 3 units apart 5 steps
Two identical positive charges, 3 units apart. Find the field and the potential at the midpoint, then decide whether a positive test charge released there would stay. This is the Two + arrangement: q₁ = +1 at (-1.5, 0), q₂ = +1 at (+1.5, 0), probe at the origin.
1. Same probe, same superposition, opposite outcome. Both charges are 1.500 from the origin and both are positive, so the two contributions are the same size and point in exactly opposite directions. They annihilate, and the field at the midpoint is 0.
  
  $E_{1} = \frac{( 1.500 , 0 )}{1. 5 ^{3}} = (0.444, 0.000), E_{2} = \frac{( - 1.500 , 0 )}{1. 5 ^{3}} = (- 0.444, 0.000) ⟹ ∣ E ∣ = 0.000$
2. The potential cannot cancel, because a scalar has no direction to cancel in. Both terms are +0.667 and V is 1.333. A point where E = 0 but V ≠ 0 is not a paradox: E is the slope of V, and a slope of zero tells you nothing about the height.
  
  $V = \frac{1}{1.500} + \frac{1}{1.500} = 0.667 + 0.667 = 1.333$
3. Zero slope is a hilltop, a valley or a mountain pass, and only one of the three holds a ball. Nudge the test charge a distance x along the line joining the charges. The nearer charge now pushes harder than the far one, the difference points back at the origin, and the restoring stiffness is 4q/a³ = 1.1852.
  
  $E_{x} (x, 0) = \frac{q}{( a + x ) ^{2}} - \frac{q}{( a - x ) ^{2}} = \frac{- 4 q a x}{( a ^{2} - x ^{2} ) ^{2}} \approx - \frac{4 q}{a ^{3}} x = - 1.1852 x$
4. Nudge it perpendicular instead. Now both charges are behind it, both push the same way, and it keeps going. Same two charges, same probe, opposite verdict — and the expelling stiffness is 2q/a³ = 0.5926, exactly half the restoring one.
  
  $E_{y} (0, y) = \frac{2 q y}{( a ^{2} + y ^{2} ) ^{3/2}} \approx + \frac{2 q}{a ^{3}} y = + 0.5926 y, E_{z} \approx + 0.5926 z$
5. Out of the plane is the perpendicular case again, because two points have nothing to distinguish one direction around their axis from another. So there are three stiffnesses, one negative and two positive, and they sum to zero. Not nearly zero.
Answer

The tool prints |E| = 0.000 from two contributions of 0.444 that cancel, and V = 1.333 from two of 0.667 that do not. The null is a pass, not a valley: stable along the line joining the charges, unstable across it. That was never a choice about where to put the charges. The three stiffnesses are the three second derivatives of V, and their sum is the divergence of E, which is zero wherever there is no charge sitting — so every direction that pulls a test charge back has to be paid for by directions that push it away. No fixed arrangement of charges can hold another charge still, however cleverly it is arranged. That is Earnshaw's theorem, and you have just obtained it from three numbers whose starting point the tool printed. Anything that really does trap a charged particle is therefore not doing it with electrostatics alone.

$\frac{\partial E _{x}}{\partial x} + \frac{\partial E _{y}}{\partial y} + \frac{\partial E _{z}}{\partial z} = - \frac{4 q}{a ^{3}} + \frac{2 q}{a ^{3}} + \frac{2 q}{a ^{3}} = - 1.1852 + 0.5926 + 0.5926 = 0$

References (1)

Insight block 3 — why the null point between two like charges is a saddle and not a trap: S. Earnshaw, "On the nature of the molecular forces which regulate the constitution of the luminiferous ether." Transactions of the Cambridge Philosophical Society 7, 97–112, 1842.

By the 42Math team · Updated 2026-07-29

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Single + - One charge at the origin: the field points straight out everywhere and the equipotentials are circles around it. The only preset with no special point.
Dipole - Along the whole line midway between them the potential is exactly 0 and the field is not - 0.500 at the centre, 0.177 two units up. Zero potential does not mean zero field.
Two + - The field is exactly 0 at the centre and the potential there is 1.3333. The mirror image of the dipole: zero field, non-zero potential.
Quadrupole - Two positives and two negatives at the corners of a square. At the centre both the field and the potential are exactly zero - the only preset here where both vanish.