Electric Field Visualiser

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

Coulomb's law: like charges repel, unlike attract	$\begin{gathered}F=\frac{k\|q_1q_2\|}{r^{2}}\\ k=\frac{1}{4\pi\varepsilon_0}=8{.}988\times10^{9}\ \mathrm{N\,m^{2}/C^{2}}\end{gathered}$
The field: force per unit charge, added as vectors	$\vec E=\frac{\vec F}{q_0}=\sum_i\frac{kq_i}{r_i^{2}}\,\hat r_i$
Potential and energy: added as plain numbers	$\begin{gathered}V=\sum_i\frac{kq_i}{r_i},\qquad W=q\,\Delta V\\ U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}\end{gathered}$

Worked example — Electric Field Visualiser

Field of q1 at P	0.25
Total charge	1
Net field at P	0.25
Its direction, from +x	0°
Potential at P	0.5

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Electric Field Visualiser

The electric field at a point is the force a small positive test charge would feel there, divided by that charge: E = F/q₀. A point charge q makes a field of size k|q|/r², pointing away from it when q is positive and towards it when q is negative, and the fields of several charges add as vectors.

What each symbol means

k: Coulomb's constant, 1/(4πε₀) = 8.988 × 10⁹ N·m²/C²
r_i: the distance from charge i to the point P
ê_i: the direction of charge i's field at P, a vector of length 1: from the charge towards P when q_i is positive, from P towards the charge when it is negative

Where the formula comes from

Coulomb's law. Two point charges push or pull along the line joining them, with a force that falls as the square of the distance.
$F=\frac{k|q_1q_2|}{r^{2}}$
Divide out the test charge. What is left belongs to the source alone. That is the field, and it is there at P whether or not anything sits at P to feel it.
$E=\frac{F}{q_0}=\frac{k|q|}{r^{2}},\qquad\vec F=q_0\vec E$
Add the fields by components. Each charge contributes its size E_i times its direction ê_i; the green arrow at P is the sum.
$\begin{gathered}E_x=\sum_i E_i\,\hat e_{i,x},\quad E_y=\sum_i E_i\,\hat e_{i,y}\\ |\vec E|=\sqrt{E_x^{2}+E_y^{2}}\end{gathered}$
Add the potentials as plain numbers. V keeps the sign of each charge and has no direction, so nothing cancels by geometry. The field points the way V falls fastest, across the dashed equipotentials.
$V=\sum_i\frac{kq_i}{r_i},\qquad\vec E=-\nabla V$
Energy. Moving a charge q from A to B takes work q(V_B − V_A) along any path. Assembling the whole arrangement from far apart takes one term for every pair.
$W=q\,(V_B-V_A),\qquad U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}$

Assumes: Charges at rest and small enough to be points. The picture is a flat slice through the charges, but the charges are ordinary three-dimensional point charges: their fields fall as 1/r² and their potentials as 1/r.

Problems solved in full

The field of one charge, and the force it exerts 3 steps
A charge of +4.0 μC sits at the origin. Find the electric field at P = (2.0 m, 0), and the force on a −2.0 μC charge placed at P.

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1. The size comes from Coulomb's law with the test charge divided out:
  
  $E=\frac{k|q|}{r^{2}}=\frac{(8{.}988\times10^{9})(4{.}0\times10^{-6})}{(2{.}0)^{2}}\approx8988\ \mathrm{N/C}$
2. A positive charge's field points away from it, along the line from the charge to P:
  
  $\hat e=+\frac{(2{.}0,\,0)}{2{.}0}=(1,\,0)$
3. The force on a charge at P is that charge times the field. A negative charge is pushed against the field:
  
  $\vec F=q_0\vec E=(-2{.}0\times10^{-6})(8988)\,\hat x\approx-1{.}798\times10^{-2}\,\hat x\ \mathrm{N}$
Answer

E = 8,988 N/C along +x; F = 1.798 × 10⁻² N along −x. The field at P is the 4.0 μC charge's alone: put a different charge at P and the force changes, the field does not.

$E\approx8988\ \mathrm{N/C}\ (+x),\qquad F\approx1{.}798\times10^{-2}\ \mathrm{N}\ (-x)$
Two charges on a line 4 steps
A charge of +2.0 μC sits at x = 0 and a charge of −3.0 μC at x = 4.0 m. Find the field at x = 1.0 m and at x = −2.0 m.

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Show solution
1. The size of each charge's field at x = 1.0 m, where the distances are 1.0 m and 3.0 m:
  
  $E_1=\frac{k\,(2{.}0\times10^{-6})}{(1{.}0)^{2}}\approx1{.}798\times10^{4},\qquad E_2=\frac{k\,(3{.}0\times10^{-6})}{(3{.}0)^{2}}\approx2996\ \mathrm{N/C}$
2. Between the charges both fields point the same way, away from the positive charge and towards the negative one, so both along +x. They add:
  
  $x=1{.}0:\quad E=E_1+E_2\approx1{.}798\times10^{4}+2996\approx2{.}097\times10^{4}\ \mathrm{N/C}\ (+x)$
3. At x = −2.0 m, to the left of both, the positive charge's field points away from it (−x) and the negative charge's towards it (+x):
  
  $x=-2{.}0:\quad E_1=\frac{k\,(2{.}0\times10^{-6})}{(2{.}0)^{2}}\approx4494\ (-x),\qquad E_2=\frac{k\,(3{.}0\times10^{-6})}{(6{.}0)^{2}}\approx749{.}0\ (+x)$
4. Opposite directions subtract, and the larger wins:
  
  $E\approx4494-749{.}0\approx3745\ \mathrm{N/C}\ (-x)$
Answer

2.097 × 10⁴ N/C along +x at x = 1.0 m; 3,745 N/C along −x at x = −2.0 m. Between two unlike charges the fields always add. Only outside them do they oppose, so that is the only place the field can vanish (problem 4).

$E(1{.}0)\approx2{.}097\times10^{4}\ \mathrm{N/C}\ (+x),\qquad E(-2{.}0)\approx3745\ \mathrm{N/C}\ (-x)$
A point off the line: add by components 4 steps
A charge of +3.0 μC sits at the origin and a charge of −2.0 μC at (0, 3.0 m). Find the size and direction of the field at P = (4.0 m, 0).

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1. The positive charge is 4.0 m away along the x axis, and its field at P points away from it, along +x:
  
  $r_1=4{.}0,\quad E_1=\frac{k\,(3{.}0\times10^{-6})}{(4{.}0)^{2}}\approx1685\ \mathrm{N/C},\quad\hat e_1=(1,\,0)$
2. The negative charge is 5.0 m away, the long side of a 3–4–5 triangle. Its field points from P back towards it: the unit vector from the charge to P is (4, −3)/5, so the field's direction is minus that:
  
  $r_2=\sqrt{4{.}0^{2}+3{.}0^{2}}=5{.}0,\quad E_2=\frac{k\,(2{.}0\times10^{-6})}{(5{.}0)^{2}}\approx719{.}0\ \mathrm{N/C},\quad\hat e_2=-\frac{(4,\,-3)}{5}=(-0{.}8,\,0{.}6)$
3. Add the components, never the sizes:
  
  $E_x=1685-0{.}8\,(719{.}0)\approx1110,\qquad E_y=0{.}6\,(719{.}0)\approx431{.}4\ \mathrm{N/C}$
4. Then rebuild the size and the angle. E_x is positive, so the inverse tangent gives the angle directly:
  
  $|\vec E|=\sqrt{1110^{2}+431{.}4^{2}}\approx1191\ \mathrm{N/C},\qquad\theta=\tan^{-1}\frac{431{.}4}{1110}\approx21{.}24^{\circ}$
Answer

1,191 N/C at 21.24° above the +x axis. Adding the two sizes would give 2,404 N/C, twice the right answer, because the two fields point 143° apart.

$|\vec E|\approx1191\ \mathrm{N/C},\qquad\theta\approx21{.}24^{\circ}$
Where the field is zero 4 steps
A charge of +4.0 μC sits at x = 0 and a charge of −1.0 μC at x = 3.0 m. Where on the x axis, other than far away, is the field zero?

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1. Two fields cancel only where they point opposite ways. Between unlike charges they point the same way (problem 2), so the point lies outside, on the side of the smaller charge, where being nearer can make up for being smaller. There the sizes must be equal:
  
  $\frac{k\,(4{.}0)}{x^{2}}=\frac{k\,(1{.}0)}{(x-3{.}0)^{2}}$
2. Cancel k and take the square root of both sides:
  
  $\frac{x^{2}}{(x-3{.}0)^{2}}=4\ \Rightarrow\ \frac{x}{x-3{.}0}=\pm2$
3. Each sign gives a candidate. At x = 2.0 m, between the charges, the sizes match but both fields point along +x, so they add; that root is rejected:
  
  $x=-2\,(x-3{.}0)\ \Rightarrow\ x=2{.}0,\qquad x=2\,(x-3{.}0)\ \Rightarrow\ x=6{.}0$
4. At x = 6.0 m the two fields are equal in size and opposite in direction:
  
  $x=6{.}0:\quad\frac{k\,(4{.}0\times10^{-6})}{(6{.}0)^{2}}\approx998{.}6\ (+x),\qquad\frac{k\,(1{.}0\times10^{-6})}{(3{.}0)^{2}}\approx998{.}6\ (-x)$
Answer

x = 6.0 m, 3.0 m beyond the −1.0 μC charge. Setting the sizes equal also finds the point where the sizes match and the directions do not, and only the directions can throw that root out.

$x=6{.}0\ \mathrm{m}$
The force on one charge of three 3 steps
Charges of +2.0 μC sit at (−3.0 m, 0) and (3.0 m, 0), and a third charge of +1.0 μC at (0, 4.0 m). Find the net force on the third charge.

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1. Find the field the other two charges make where the third one sits. Both are 5.0 m away:
  
  $r=\sqrt{3{.}0^{2}+4{.}0^{2}}=5{.}0,\qquad E_1=E_2=\frac{k\,(2{.}0\times10^{-6})}{(5{.}0)^{2}}\approx719{.}0\ \mathrm{N/C}$
2. By symmetry the x components cancel and the y components add:
  
  $\hat e_1=\frac{(3,\,4)}{5},\quad\hat e_2=\frac{(-3,\,4)}{5}\ \Rightarrow\ E_x=0,\quad E_y=2\,(0{.}8)(719{.}0)\approx1150\ \mathrm{N/C}$
3. The force on the third charge is its charge times that field, along +y, away from the pair:
  
  $F=q_3E=(1{.}0\times10^{-6})(1150)\approx1{.}15\times10^{-3}\ \mathrm{N}$
Answer

1.15 × 10⁻³ N along +y. The third charge is left out of the field at its own position: a charge's field acts on every other charge and never on itself.

$\vec F\approx1{.}15\times10^{-3}\ \mathrm{N}\ (+y)$
Potential, and the work to move a charge 3 steps
The two +2.0 μC charges of problem 5 stay where they are, and the +1.0 μC charge is moved from A = (0, 4.0 m) to the origin B. How much work does the mover do?

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1. Potential is a plain sum with no directions. At A both charges are 5.0 m away:
  
  $V_A=\frac{k\,(2{.}0\times10^{-6})}{5{.}0}+\frac{k\,(2{.}0\times10^{-6})}{5{.}0}\approx7190\ \mathrm{V}$
2. At B both are 3.0 m away:
  
  $V_B=\frac{k\,(2{.}0\times10^{-6})}{3{.}0}+\frac{k\,(2{.}0\times10^{-6})}{3{.}0}\approx1{.}198\times10^{4}\ \mathrm{V}$
3. The work done against the field is the charge times the rise in potential, whatever path it takes:
  
  $W=q\,(V_B-V_A)\approx(1{.}0\times10^{-6})(11983-7190)\approx4{.}793\times10^{-3}\ \mathrm{J}$
Answer

4.793 × 10⁻³ J. The field at B is zero, because the two pushes cancel there, and the work to get there is not: work follows the potential, which rises all the way from A to B, not the field at the end point.

$W\approx4{.}793\times10^{-3}\ \mathrm{J}$
The energy of an arrangement 3 steps
Charges of +3.0 μC at the origin, −2.0 μC at (0, 3.0 m) and +1.0 μC at (4.0 m, 0) are brought together from far apart. How much work did that take?

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1. Bring them in one at a time. The first costs nothing; each later one costs its charge times the potential the earlier ones make where it lands. Added up, every pair appears once:
  
  $U=\sum_{i<j}\frac{kq_iq_j}{r_{ij}}=k\Bigl(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\Bigr)$
2. The three distances:
  
  $r_{12}=3{.}0,\qquad r_{13}=4{.}0,\qquad r_{23}=\sqrt{3{.}0^{2}+4{.}0^{2}}=5{.}0$
3. Each pair keeps its signs:
  
  $U=(8{.}988\times10^{9})\Bigl(\frac{(3)(-2)}{3}+\frac{(3)(1)}{4}+\frac{(-2)(1)}{5}\Bigr)\times10^{-12}=(8{.}988\times10^{-3})(-1{.}65)\approx-1{.}483\times10^{-2}\ \mathrm{J}$
Answer

−1.483 × 10⁻² J. Negative: the two attracting pairs outweigh the one repelling pair, so the field did the work, and pulling the three apart again would cost 1.483 × 10⁻² J.

$U\approx-1{.}483\times10^{-2}\ \mathrm{J}$
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.

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Show solution
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.
  
  $\vec E(\vec p)=\sum_i q_i\,\frac{\vec p-\vec x_i}{|\vec p-\vec x_i|^{3}}, \qquad \vec p=(0,0)$
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.
  
  $q_1=+1,\ \vec x_1=(-2,0):\quad \vec r_1=(2,\,0),\ r_1=2, \quad \vec E_1=\frac{+1\cdot(2,0)}{2^{3}}=(0{.}25,\,0)$
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.
  
  $q_2=-1,\ \vec x_2=(2,0):\quad \vec r_2=(-2,\,0),\ r_2=2, \quad \vec E_2=\frac{-1\cdot(-2,0)}{2^{3}}=(0{.}25,\,0)$
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.
  
  $\vec E=\vec E_1+\vec E_2=(0{.}5,\,0),\qquad |\vec E|=0{.}5$
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.
  
  $V=\sum_i\frac{q_i}{r_i}=\frac{+1}{2}+\frac{-1}{2} =0{.}5-0{.}5=0$
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.
  
  $\vec p=(0,y):\quad E_y=0,\qquad E_x=\frac{2qa}{(a^{2}+y^{2})^{3/2}},\ a=2;\qquad y=0:\ \frac{4}{8}=0{.}5$
Answer

The tool prints |E| = 0.5 at the origin, assembled from two contributions of 0.25 that point the same way, and V = 0 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. 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.

$y\gg a:\ \ E_x\to\frac{2qa}{y^{3}}=\frac{4}{y^{3}}, \qquad \frac{E_x}{q/y^{2}}=\frac{2a}{y}=\frac{4}{y};\qquad y=20:\ 0.2$
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.

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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.
  
  $\vec E_1=\frac{(1{.}5,\,0)}{1{.}5^{3}}=(0{.}444,\,0),\quad \vec E_2=\frac{(-1{.}5,\,0)}{1{.}5^{3}}=(-0{.}444,\,0) \ \Longrightarrow\ |\vec E|=0$
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.
  
  $V=\frac{1}{1{.}5}+\frac{1}{1{.}5}=0{.}667+0{.}667=1{.}333$
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.
  
  $E_x(x,0)=\frac{q}{(a+x)^{2}}-\frac{q}{(a-x)^{2}} =\frac{-4qax}{(a^{2}-x^{2})^{2}}\ \approx\ -\frac{4q}{a^{3}}\,x=-1.1852\,x$
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.
  
  $E_y(0,y)=\frac{2qy}{(a^{2}+y^{2})^{3/2}} \ \approx\ +\frac{2q}{a^{3}}\,y=+0.5926\,y,\qquad E_z\approx+0.5926\,z$
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 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 minus the three second derivatives of V, and their sum is the divergence of E, which is zero wherever no charge sits, 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 in stable equilibrium, 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. A trap that holds a charged particle at rest therefore needs more than a static electric field: a Penning trap adds a magnetic field, a Paul trap makes the electric field oscillate.

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

Supplementary problems, answers only

Find the size and direction of the field 30 cm from a charge of −4.0 nC.

Show answer

399.4 N/C, pointing towards the charge.

$E=\frac{k|q|}{r^{2}}\approx399{.}4\ \mathrm{N/C}$
Charges of +4.0 μC at x = 0 and +1.0 μC at x = 3.0 m. Where between them is the field zero?

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x = 2.0 m.

$x=2{.}0\ \mathrm{m}$
Charges of +1.0 μC sit at (−1.0 m, 0) and (1.0 m, 0). Find the field at (0, 1.0 m).

Show answer

6,355 N/C along +y.

$E=2\cdot\frac{k\,(1{.}0\times10^{-6})}{2}\cdot\frac{1}{\sqrt2}\approx6355\ \mathrm{N/C}\ (+y)$
A square of side 2.0 m has +1.0 μC at three corners and −1.0 μC at the fourth. Find the field and the potential at its centre.

Show answer

8,988 N/C, pointing at the negative corner; 1.271 × 10⁴ V.

$E\approx8988\ \mathrm{N/C},\qquad V=\frac{2kq}{\sqrt2}\approx1{.}271\times10^{4}\ \mathrm{V}$
A 2.0 g bead carrying +3.0 μC is released from rest 0.20 m from a fixed +5.0 μC charge. How fast is it moving when it is far away?

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25.96 m/s: all of the 0.6741 J of energy becomes kinetic energy.

$\tfrac12mv^{2}=\frac{kq_1q_2}{r}\approx0{.}6741\ \mathrm{J},\qquad v\approx25{.}96\ \mathrm{m/s}$
Four charges of +1.0 μC sit at the corners of a square of side 1.0 m. How much work did it take to put them there?

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4.866 × 10⁻² J: four sides and two diagonals, (4 + √2)kq²/a.

$U=\bigl(4+\sqrt2\bigr)\frac{kq^{2}}{a}\approx4{.}866\times10^{-2}\ \mathrm{J}$

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.

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.

Learning path

Where fields come from

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.