Electric field intensity

Coulomb's law tells you the force between two charges, but it needs both of them present at once — awkward if you want to describe the space around a single charge before any second charge shows up. Electric field intensity fixes that: it packages "how strongly would a charge be pushed or pulled here" into a property of the point itself, independent of whatever test charge you eventually bring in. This idea — force per unit test charge, $\vec E = \vec F/q_0$ — turns Coulomb's law inside out. Once you know $\vec E$ at a point, the force on any charge dropped there is just $q_0\vec E$ , no re-derivation needed. It also opens the door to superposition (adding fields from several sources), field lines (a picture of the field), and — later in this ISC chapter — Gauss's theorem, which makes finding $\vec E$ for symmetric charge distributions almost effortless.

The core insight

Electric field intensity $\vec E = \vec F/q_0$ is the force per unit positive test charge at a point, equal to $kq/r^2$ directed radially from a point charge $q$ , and net fields from multiple sources add vectorially by superposition.

The core idea

Definition. The electric field intensity at a point is the force a small positive test charge would feel there, divided by the size of that test charge:

\vec E = \frac{\vec F}{q_0}

where: $\vec F$ is the electrostatic force on the test charge (newton, N), $q_0$ is the magnitude of the test charge (coulomb, C), and $\vec E$ is the field (newton per coulomb, N/C, equivalent to volt per metre, V/m). The test charge is taken vanishingly small precisely so it does not itself disturb the source charges producing the field being measured — you are probing the field, not changing it.

Field of a point charge. Substitute Coulomb's law for $\vec F$ and the definition collapses to a formula that depends only on the source charge and distance, not on $q_0$ at all:

E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} = k\frac{q}{r^2}

where: $q$ is the source charge, $r$ is the distance from it to the point, and $k = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ . The direction is radially outward if $q$ is positive, radially inward (toward $q$ ) if $q$ is negative. This independence from $q_0$ is the whole point of the definition — $\vec E$ belongs to the source and the location, not to whatever you use to measure it.

Superposition. When several charges act together, each produces its own field as if the others were not there, and the net field is the vector sum:

\vec E_{\text{net}} = \vec E_1 + \vec E_2 + \vec E_3 + \cdots

This is the single most-used tool in the chapter: break a messy charge arrangement into point charges, find each field, add as vectors (components if not collinear).

Continuous distributions (conceptually). A charged rod, disc, or sphere can be sliced into infinitesimal elements $dq$ , each contributing $d\vec E = k\,dq/r^2$ at the observation point; summing (integrating) over the whole object gives the net field. This ISC syllabus leans on Gauss's theorem — covered later in the chapter — for the standard symmetric cases rather than direct integration, but the underlying idea is the same superposition principle applied to infinitely many infinitesimal point charges.

Figure

Field vector E at point P pointing radially away from a positive point charge q at distance r

Derivation AElectric field of a single point charge, from the definition and Coulomb's law

Starting from the definition of electric field intensity as force per unit test charge, and combining it with Coulomb's law for the force between two point charges, derive the formula that gives the field of an isolated point charge $q$ at any distance $r$ , showing explicitly why the result does not depend on the test charge used to measure it.

Step 1. Place a small positive test charge $q_0$ at a point that is a distance $r$ from a source point charge $q$ . By Coulomb's law, the force on $q_0$ has magnitude

F = \frac{1}{4\pi\varepsilon_0}\frac{|q\,q_0|}{r^2} = k\frac{|q|\,q_0}{r^2}

directed along the line joining the two charges — away from $q$ if both are positive (repulsion), toward $q$ if $q$ is negative and $q_0$ positive (attraction).

Step 2. Apply the definition $\vec E = \vec F/q_0$ : divide the force by the test charge $q_0$ .

E = \frac{F}{q_0} = \frac{1}{4\pi\varepsilon_0}\frac{|q|\,q_0}{r^2\,q_0} = \frac{1}{4\pi\varepsilon_0}\frac{|q|}{r^2} = k\frac{|q|}{r^2}

The test charge $q_0$ cancels out completely. This cancellation is not a coincidence — it is guaranteed because the Coulomb force is proportional to $q_0$ , so dividing by $q_0$ always removes it, leaving an expression that depends only on the source charge $q$ and the distance $r$ .

Step 3. Restore the sign and direction. If $q$ is positive, the force on any positive test charge points away from $q$ , so $\vec E$ points radially outward. If $q$ is negative, the force on a positive test charge points toward $q$ , so $\vec E$ points radially inward. The magnitude in both cases is $E = k\,|q|/r^2$ , and this is the field a lone point charge sets up at every point of the space around it — a value you can now use with $\vec F = q_0\vec E$ to predict the force on any charge placed there, without repeating the Coulomb's-law calculation each time.

Worked example

Find the electric field intensity $50\ \text{cm}$ from a point charge of $+2\ \mu\text{C}$ in air, the force it would exert on a $-3\ \mu\text{C}$ charge placed there, and then the net field at the midpoint between this charge and an equal-and-opposite $-2\ \mu\text{C}$ charge placed $40\ \text{cm}$ away from it.

Step 1. Field due to the single charge. Given $q = 2\times10^{-6}\ \text{C}$ and $r = 0.50\ \text{m}$ :

E = k\frac{q}{r^2} = (9\times10^9)\times\frac{2\times10^{-6}}{(0.50)^2} = (9\times10^9)\times\frac{2\times10^{-6}}{0.25} = 7.2\times10^4\ \text{N/C}

directed radially away from the $+2\ \mu\text{C}$ charge, since it is positive.

Step 2. Force on a test charge placed there. For $q_0 = -3\times10^{-6}\ \text{C}$ at that same point, use $F = q_0 E$ :

F = (3\times10^{-6})\times(7.2\times10^4) = 0.216\ \text{N}

Since $q_0$ is negative, this force points opposite to $\vec E$ — toward the $+2\ \mu\text{C}$ charge, consistent with attraction between unlike charges.

Step 3. Now place a second charge $q_2 = -2\ \mu\text{C}$ so that, together with the original $q_1 = +2\ \mu\text{C}$ scaled up conceptually to the earlier dipole pair of magnitude $5\ \mu\text{C}$ each $40\ \text{cm}$ apart (as commonly set in this problem type), find the field at the midpoint M, $r = 0.20\ \text{m}$ from each charge. Field due to $q_1 = +5\ \mu\text{C}$ at M points away from it, from the positive charge toward the negative one:

E_1 = k\frac{q_1}{r^2} = (9\times10^9)\times\frac{5\times10^{-6}}{(0.20)^2} = 1.125\times10^6\ \text{N/C}

Field due to $q_2 = -5\ \mu\text{C}$ at M points toward it — the same direction, since M lies between the two charges:

E_2 = k\frac{|q_2|}{r^2} = 1.125\times10^6\ \text{N/C}

Step 4. Apply superposition. Both fields point the same way at the midpoint, so they add rather than cancel:

E_{\text{net}} = E_1 + E_2 = 1.125\times10^6 + 1.125\times10^6 = \mathbf{2.25\times10^6\ \text{N/C}}

directed from the positive charge toward the negative charge — the signature result for the midpoint of a dipole axis, and a reminder that "opposite charges" does not mean "cancelling fields" at every point.

Competitive stretch

If the midpoint were instead moved off the axis — say to a point equidistant from both charges but perpendicular to the line joining them — the two field vectors would no longer be collinear. You would then resolve $E_1$ and $E_2$ into components along and perpendicular to the axis before adding; the components along the axis reinforce (as above) while the perpendicular components cancel by symmetry. Recognising which geometry lets you add magnitudes directly, and which forces component-wise addition, is exactly what separates a fast, correct solution from a slow, error-prone one on this topic.

Common traps

Flipping the direction convention: field due to a positive charge points away from it; field due to a negative charge points toward it. Getting this backward is the single most common sign error in this topic.
Treating $\vec E$ as if it depended on the test charge $q_0$ — it does not; $q_0$ cancels in the derivation and $\vec E$ is fixed entirely by the source charges and the point in space.
Adding field magnitudes arithmetically when the individual vectors are not collinear — vector addition (components) is required unless geometry guarantees the fields lie along one line.
Treating N/C and V/m as two different quantities rather than the same unit expressed two ways.

Ask the tutor: if you want more practice with off-axis superposition problems that require component-wise vector addition, ask the AI tutor for extra worked problems and a step-by-step method.

Why does the electric field not depend on the test charge used to measure it

Because the Coulomb force on the test charge is itself proportional to the test charge's own magnitude, so when you divide force by test charge in the definition $\vec E = \vec F/q_0$ , that dependence cancels exactly. What is left, $E = kq/r^2$ , depends only on the source charge $q$ and the distance $r$ — properties of the source and the point, not of whoever is measuring. This is why field is a useful concept: compute it once, then find the force on any charge you place there.

If two equal and opposite charges are placed symmetrically, does the field always cancel at the midpoint

No — this is a common misconception. At the midpoint of a dipole (equal and opposite charges), the two fields point in the same direction, from the positive charge toward the negative one, so they add. Fields from a positive and a negative charge only cancel at the midpoint when both charges have the same sign and equal magnitude, because then both fields point away from (or both toward) the charges in opposite directions along the line, and the equal magnitudes cancel exactly.

Is electric field intensity the same thing as electric potential

No, though they are closely related and often confused. Electric field intensity $\vec E$ is a vector (force per unit charge, N/C), describing how strongly and in which direction a charge would be pushed at a point. Electric potential is a scalar (energy per unit charge, measured in volts), describing how much work is needed to bring a charge there from infinity. The field is, in fact, related to how fast the potential changes with position — but that connection is developed later in the chapter and is not needed to compute $\vec E = kq/r^2$ here.

Electric field intensity

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads