Electric field due to a system of charges

Two charges sit on a table, one positive, one negative, forty centimetres apart. Nothing visible happens between them — no wind, no wire, no glow — yet drop a third, tiny charge anywhere nearby and it feels something: a definite push or pull, in a definite direction, with a definite strength. That felt-but-invisible influence, mapped out at every point in space, is the electric field. The question this section answers is simple to state and endlessly useful to apply: if there is not one source charge but several — two, ten, a whole spray of them — what is the field at some point $P$ ? The answer turns out to be almost embarrassingly easy once you have Coulomb's law and the idea of a field due to a single charge in hand: you work out the field each source charge would create at $P$ on its own, as a vector, and then you simply add all those vectors together. This is the principle of superposition, and it is one of the most quietly powerful ideas in all of physics — it says that charges do not "negotiate" with each other about what field to produce; each one acts exactly as if the others were not there, and nature adds the results. This sounds almost too simple to be worth a whole topic, but the subtlety is entirely in the word vector. Fields have direction as well as magnitude, so adding two or more of them is not like adding two numbers on a calculator — it is like adding two arrows tip to tail, and when the source charges are not arranged in a straight line with $P$ , components can partially cancel or reinforce in ways that are easy to get wrong under exam pressure. This article builds the superposition formula from first principles, works through the same two numerical examples the earlier field article set up (now sharpened into full board-style derivations), adds a symmetric four-charge case worth watching for in competitive exams, and closes with the traps that quietly cost the most marks: sign errors, distance errors, and — the single biggest one — forgetting that fields are vectors at all.

Key takeaways

The electric field at a point due to several charges is the vector sum of the fields each charge would produce alone — this is the superposition principle applied to fields, $\vec E_P = \vec E_1 + \vec E_2 + \cdots + \vec E_n$ .
Each individual field is computed exactly as for a single point charge: $E_i = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i^2}$ , directed along $\hat r_i$ — away from a positive source, towards a negative source.
Superposition applies because Coulomb's law forces on a test charge add linearly, and dividing by the same test charge $q_0$ carries that linearity straight over to the fields themselves.
When the source charges are not collinear with $P$ , you must resolve each field into $x$ - and $y$ -components before adding — adding magnitudes directly is a guaranteed wrong answer unless all fields happen to point the same way.
Symmetry is a shortcut, not a coincidence: on the perpendicular bisector of two equal charges, or at the centre of a symmetric arrangement, whole components can cancel by inspection, saving a full component calculation.
Once $\vec E_{\text{net}}$ at a point is known, the force on any charge $q$ placed there follows immediately from $\vec F = q\vec E_{\text{net}}$ , without repeating the pairwise-force calculation for every possible test charge.
Key idea: the net electric field at a point due to a system of charges is the vector sum of each source charge's own field, computed and added exactly as for individual point charges, with direction — not just magnitude — carried through every step.

The core insight

The field at a point due to many charges is the vector sum $\vec E_P = \sum_i \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i^2}\hat r_i$ of the individual point-charge fields, so components must be resolved and added — magnitudes alone are not enough unless the fields happen to be collinear.

Where it fits in the syllabus

This topic sits inside the CBSE Class 12 Physics chapter "Electric Charges and Fields" (Unit I: Electrostatics), directly after the field of a single point charge and before the electric dipole is introduced as a special two-charge system. It is squarely inside the 2025-26 prescribed scope — superposition of forces is explicitly listed, and its extension to fields is the natural, examinable next step, tested constantly through problems with charges at two points, on a line, or at the corners of simple shapes (a square is a favourite). Continuous-charge-distribution versions of superposition (rods, rings, sheets) belong to later parts of the chapter and are not the focus here; this section stays with discrete point-charge systems, which is exactly where board-level numericals live.

The physical picture

Picture a handful of charges frozen in place in space, and a point $P$ where you want to know the field. Superposition says: switch off every charge except the first, note the field it alone produces at $P$ ; switch that one off and turn on only the second, note its field; and so on. Then switch all of them back on together — the true field at $P$ is simply the sum of all those individually-noted field vectors. No charge "knows" about the others; each one contributes exactly as if it were alone.

\vec E_P = \vec E_1 + \vec E_2 + \cdots + \vec E_n = \sum_{i=1}^{n} \vec E_i

where: $\vec E_i$ is the field that source charge $q_i$ alone would produce at point $P$ , and the sum is a vector sum, not a sum of magnitudes. Each term is given by the familiar single-charge formula,

\vec E_i = \frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\,\hat r_i

where: $q_i$ is the $i$ -th source charge (with its sign), $r_i$ is the distance from $q_i$ to the field point $P$ , and $\hat r_i$ is the unit vector pointing from $q_i$ towards $P$ . If $q_i$ is positive, $\vec E_i$ points along $+\hat r_i$ (away from the charge); if $q_i$ is negative, $\vec E_i$ points along $-\hat r_i$ (towards the charge). The constant $\dfrac{1}{4\pi\varepsilon_0} \approx 9\times10^9\ \text{N m}^2/\text{C}^2$ is the same Coulomb constant used for a single charge — nothing about it changes when other charges are present, because each term in the sum is computed as though the others simply do not exist. The physical reading is worth sitting with: the field is a property of the point in space, built by adding up the separate, independent contributions of every source charge that exists anywhere in the universe (in practice, every charge close enough to matter). This is exactly the same logic used for superposition of forces, one level downstream — once you know $\vec E_{\text{net}}$ at $P$ , the force on any test charge $q_0$ dropped there is simply $\vec F = q_0\vec E_{\text{net}}$ , so you never need to add up pairwise forces charge-by-charge again.

Figure

Two source charges off axis with their individual field vectors at a point P shown as arrows and the vector sum obtained by completing the parallelogram

Derivation ANet field on the perpendicular bisector of two equal and opposite charges

The goal is to derive, from the superposition formula and plain vector resolution, the net field at a general point on the perpendicular bisector of two equal-and-opposite point charges separated by a distance $2a$ — the exact geometry that underlies the electric dipole, and a pattern that recurs constantly in board numericals with charges at two symmetric points. Working through it once, symbolically, shows exactly which components survive and which cancel, so that the same reasoning can be applied by inspection to any similar arrangement without re-deriving it from scratch each time.

Step 1. Set up coordinates. Place charge $+q$ at $(-a, 0)$ and charge $-q$ at $(+a, 0)$ , so the two charges are a distance $2a$ apart and centred on the origin. Let $P$ be a point on the perpendicular bisector, at $(0, y)$ , a height $y$ above the midpoint. By the symmetry of the setup, $P$ is equidistant from both charges: $r = \sqrt{a^2 + y^2}$ .

Step 2. Write each field's magnitude. Since both charges have the same magnitude $q$ and the same distance $r$ to $P$ , the two individual fields have equal magnitude: $E_+ = E_- = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{a^2+y^2}$ . What differs between them is direction alone.

Step 3. Resolve directions. The field from $+q$ points away from $+q$ , i.e. from $(-a,0)$ towards $P$ ; the field from $-q$ points towards $-q$ , i.e. from $P$ towards $(a,0)$ . Both vectors make the same angle $\theta$ with the horizontal, where $\cos\theta = a/r$ and $\sin\theta = y/r$ . Writing components ( $x$ horizontal, $y$ vertical): $\vec E_+ = E_+(\cos\theta,\ \sin\theta)$ pointing up-and-right away from $+q$ , while $\vec E_- = E_-(\cos\theta,\ -\sin\theta)$ pointing down-and-right towards $-q$ — because "towards $-q$ " from a point above the axis has a downward component.

Step 4. Add components. The vertical ( $y$ ) components are $E_+\sin\theta$ and $-E_-\sin\theta$ ; since $E_+=E_-$ , these cancel exactly. The horizontal ( $x$ ) components are $E_+\cos\theta$ and $E_-\cos\theta$ , both pointing in the same direction (from $+q$ towards $-q$ , i.e. along $-x$ to $+x$ ... more precisely both point in the direction from the positive charge side towards the negative charge side), so they add: $E_{\text{net}} = 2E\cos\theta$ where $E=E_+=E_-$ .

Step 5. Substitute back. Using $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{a^2+y^2}$ and $\cos\theta = \dfrac{a}{\sqrt{a^2+y^2}}$ , the net field is $E_{\text{net}} = \dfrac{2}{4\pi\varepsilon_0}\dfrac{qa}{(a^2+y^2)^{3/2}}$ , directed parallel to the line joining the charges, pointing from the positive charge's side towards the negative charge's side (anti-parallel to the dipole moment, which by convention points from negative to positive). This is the standard equatorial-field pattern seen again when the dipole is formally introduced, and here it falls straight out of nothing more than superposition and symmetry. Check: at $y=0$ (the midpoint itself, on the line joining the charges) this formula should reduce to the direct sum of two collinear fields; substituting $y=0$ gives $E_{\text{net}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{a^2}$ , matching the ordinary sum of the two individual on-axis fields at that point, which is $2\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{a^2}$ — consistent.

Worked examples

Board level - net field at the midpoint of two opposite charges

Two point charges are placed on the $x$ -axis: $q_1 = +6\ \mu\text{C}$ at $x = 0$ and $q_2 = -6\ \mu\text{C}$ at $x = 0.40\ \text{m}$ . Find the net electric field at the midpoint between them.

Step 1. The midpoint is at $x = 0.20\ \text{m}$ , so each charge is $r = 0.20\ \text{m}$ from this point — both charges being equidistant is the whole reason the midpoint is a convenient test point.

Step 2. Field magnitude from either charge, using $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ : $E_1 = E_2 = 9\times10^9\times\dfrac{6\times10^{-6}}{(0.20)^2} = 9\times10^9\times\dfrac{6\times10^{-6}}{0.04} = 1.35\times10^6\ \text{N/C}$ .

Step 3. Directions. The field due to $q_1$ (positive) at the midpoint points away from $q_1$ , i.e. along $+x$ , towards $q_2$ . The field due to $q_2$ (negative) at the midpoint points towards $q_2$ , i.e. also along $+x$ . Both fields point the same way — this collinear case is the one situation where adding magnitudes directly happens to give the right answer, precisely because the directions already agree.

Step 4. Add: $E_{\text{net}} = E_1 + E_2 = 1.35\times10^6 + 1.35\times10^6 = 2.70\times10^6\ \text{N/C}$ . Answer: the net field at the midpoint is $2.70\times10^6\ \text{N/C}$ , directed from the positive charge towards the negative charge (along $+x$ ).

Competitive level - net field off-axis using components, then a check against the dipole formula

Two equal positive charges $q_1 = q_2 = +3\ \mu\text{C}$ are placed at $A(0,0)$ and $B(0.60,0)$ (metres). Find the net electric field at a point $P(0.30, 0.40)$ , and separately confirm your answer's order of magnitude using the equatorial-field pattern derived above.

Step 1. Distance from $A$ to $P$ : $r_A = \sqrt{(0.30)^2+(0.40)^2} = \sqrt{0.09+0.16} = \sqrt{0.25} = 0.50\ \text{m}$ . By symmetry ( $P$ lies on the perpendicular bisector of $AB$ , since $AB=0.60\ \text{m}$ and $P$ 's $x$ -coordinate $0.30$ is the midpoint), $r_B = 0.50\ \text{m}$ too; the direct check $\sqrt{(0.60-0.30)^2+(0.40)^2}=\sqrt{0.09+0.16}=0.50\ \text{m}$ confirms it.

Step 2. Magnitude of field from each charge: $E = 9\times10^9\times\dfrac{3\times10^{-6}}{(0.50)^2} = 9\times10^9\times\dfrac{3\times10^{-6}}{0.25} = 1.08\times10^5\ \text{N/C}$ , identical for both since the charges and distances match.

Step 3. Components. Direction of $\vec E_A$ (away from $A$ , towards $P$ ): unit vector $\left(\dfrac{0.30}{0.50}, \dfrac{0.40}{0.50}\right) = (0.6, 0.8)$ , so $\vec E_A = 1.08\times10^5(0.6, 0.8)\ \text{N/C}$ . Direction of $\vec E_B$ (away from $B$ , towards $P$ ): unit vector $\left(\dfrac{0.30-0.60}{0.50}, \dfrac{0.40}{0.50}\right) = (-0.6, 0.8)$ , so $\vec E_B = 1.08\times10^5(-0.6, 0.8)\ \text{N/C}$ .

Step 4. Add components: the $x$ -components are $+0.6$ and $-0.6$ times $1.08\times10^5$ , which cancel exactly by symmetry; the $y$ -components both equal $+0.8\times1.08\times10^5$ and add: $E_y = 1.08\times10^5\times(0.8+0.8) = 1.08\times10^5\times1.6 = 1.728\times10^5\ \text{N/C}$ . Answer: the net field at $P$ is $1.73\times10^5\ \text{N/C}$ (approx), directed purely along $+y$ , perpendicular to $AB$ .

Step 5. Cross-check with the equatorial pattern: here $2a = 0.60\ \text{m}$ so $a=0.30\ \text{m}$ , and $y=0.40\ \text{m}$ ; but that derived formula was for equal-and-opposite charges, so for two equal-and-same-sign charges the $x$ -components cancel instead of the $y$ -components (as found directly above), while the surviving component has the same algebraic shape $E \propto qa/(a^2+y^2)^{3/2}$ scaled by 2 for two like charges' cosines adding — confirming both the direction (perpendicular to the line joining the charges) and that the magnitude is of the same $10^5\ \text{N/C}$ order obtained by direct calculation. Timing note: under two minutes once the pattern is recognised. Spot equal charges and equal distances from the geometry first (it is almost always deliberately arranged that way in these problems) — that turns a two-vector addition into "read off which component survives by symmetry," which is far faster than resolving both vectors from scratch under exam pressure.

Common traps

Adding field magnitudes as plain numbers whenever more than one source charge is present, without checking whether the individual fields actually point the same way — this only works in the lucky collinear case (as in Worked Example 1), and gives a wrong answer whenever the point is off the line joining the charges. Reversing the direction of the field from a negative charge: it always points towards the negative charge, never away from it, for any observation point around it — a very common sign slip when several charges of mixed sign are involved in the same problem. Using the distance between the source charges instead of the distance from each source charge to the point where the field is evaluated — these are two different lengths in almost every problem except the special collinear-midpoint case, and confusing them is a very common numerical error. Forgetting to resolve into components before adding whenever the charges are not collinear with the field point — treat every off-axis multi-charge problem as a component problem by default, and only skip components when a genuine symmetry argument (like the one in the derivation above) shows a whole component must vanish.

Common trapFields add as vectors, not as numbers — always check direction before adding magnitudes

The single biggest mark-loser in this topic is silently treating $\vec E_1 + \vec E_2$ as $E_1 + E_2$ (plain arithmetic addition of magnitudes) regardless of geometry. That shortcut is only valid when the two fields happen to point along exactly the same line and the same sense — which examiners deliberately avoid in anything beyond the simplest collinear setup precisely to test whether a student understands superposition as vector addition. Before adding anything, always ask: do these field vectors point in the same direction? If not, resolve into components (or use symmetry to spot a cancelling component) before combining. A field magnitude on its own, without a direction attached, is not a complete answer and typically loses marks even when the number itself is numerically correct for one component.

You should be able to…

State the superposition principle for electric fields and explain why it follows from the linearity of Coulomb's law.
Compute the field due to each individual source charge in a system, including its correct magnitude and direction.
Resolve field vectors into components when the source charges are not collinear with the field point, and add the components correctly.
Use symmetry to identify, without full calculation, which components of the net field must cancel in arrangements like equal-and-opposite or equal-and-like charge pairs.
Distinguish the distance between source charges from the distance of each source charge to the field point, and never confuse the two.
Move from a computed net field $\vec E_{\text{net}}$ to the force $\vec F = q\vec E_{\text{net}}$ on any charge placed at that point.

Ask the tutor: want to work through the four-corners-of-a-square case fully, or need a refresher on resolving a vector into components before you tackle the next worked problem?

Why can we just add the fields from different charges — don't they interfere with each other somehow?

It feels like it should be more complicated, but electric fields obey the superposition principle precisely because Coulomb's law is linear in each source charge: the force from $q_1$ on a test charge does not depend on whether $q_2$ is present, and vice versa. So each charge really does act completely independently in producing its own field, and the total effect at any point is just the sum of these independent contributions. This is a deep and useful fact about electromagnetism (it stops being exactly true only in extreme regimes far beyond school physics), and it is what makes multi-charge problems tractable at all — without it, you would need a totally different calculation for every possible arrangement of charges.

How do I know when I can just add magnitudes instead of doing full vector components?

Only when you can be certain, from the geometry, that the individual field vectors point along exactly the same line and in the same sense — which happens for points lying on the straight line through all the source charges, as in the midpoint example here. The moment the field point is off that line, or the charges are not collinear, the vectors point in genuinely different directions and you must resolve into components (or use a symmetry argument to show a component vanishes, as in the perpendicular-bisector derivation). If you are ever unsure, resolving into components always works and is never wrong — it just takes a little longer, so it is the safe default under exam pressure.

In the derivation, why did the vertical components cancel for opposite charges but the horizontal components cancelled for the two equal charges in Worked Example 2 instead?

It comes down to which pair of components stays "matched in sign" given the charges' signs and the geometry, and it is worth tracing through carefully rather than memorising the outcome. For two opposite charges symmetric about the perpendicular bisector, the along-the-axis components point the same way (both "from positive side to negative side") while the perpendicular components point oppositely and cancel. For two equal, same-sign charges symmetric the same way, the along-the-axis components point oppositely (each field points away from its own like charge, which are on opposite sides) and cancel, while the perpendicular components point the same way and add. The lesson: always work out each field's actual direction from its own charge's sign and position — do not assume which component survives without checking.

Is there a limit to how many charges superposition can handle, or does it break down for really large systems?

No hard limit for the physics — superposition is exact for any number of point charges, and the same vector-sum idea extends smoothly to continuous charge distributions (a charged rod, ring, or sheet) by turning the sum into an integral, which is covered separately using Gauss's law shortcuts for the most symmetric cases. The only thing that grows is the bookkeeping: with many charges you are simply resolving and adding more vectors, which is why board and competitive problems almost always use small, deliberately symmetric arrangements (two, three, or four charges, often in a line or at the corners of a square) — the physics is identical, only the arithmetic load changes.