Electric Field Due to a System of Charges and Its Physical Significance

So far, the electric field has been defined for a single point charge sitting alone in space. But real charge distributions almost always involve many charges acting together. How do you find the net electric field when several charges are all contributing at once? The answer relies on the same superposition principle that works for forces. And once that is in hand, a deeper question arises: is the electric field just a mathematical convenience, or is it something physically real?

Finding the Net Field: Superposition Applied to Electric Fields

Imagine a collection of charges $q_1, q_2, \ldots, q_n$ located at position vectors $\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_n$ relative to some origin O. You want to know the electric field at some point P (position vector $\mathbf{r}$) in the space around them.

The core idea is simple: each charge produces its own electric field at P, completely independently of every other charge. The net field at P is the vector sum of all these individual contributions.

Building the Result One Charge at a Time

The field at P due to $q_1$ alone (located at $\mathbf{r}_1$) is:

$\mathbf{E}_1 = \frac{1}{4\pi\varepsilon_0} \frac{q_1}{r_{1P}^2} \hat{\mathbf{r}}_{1P}$

Here, $r_{1P}$ is the distance from $q_1$ to P, and $\hat{\mathbf{r}}_{1P}$ is a unit vector (a vector with magnitude 1) pointing from $q_1$ toward P.

Similarly, the field at P due to $q_2$ at $\mathbf{r}_2$ is:

$\mathbf{E}_2 = \frac{1}{4\pi\varepsilon_0} \frac{q_2}{r_{2P}^2} \hat{\mathbf{r}}_{2P}$

You can write an identical expression for every charge in the system: $\mathbf{E}_3, \mathbf{E}_4, \ldots, \mathbf{E}_n$.

Fig 1.9: Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges

The General Superposition Formula

By the superposition principle, the total electric field at P is the sum of all individual fields:

$\mathbf{E}(\mathbf{r}) = \mathbf{E}_1(\mathbf{r}) + \mathbf{E}_2(\mathbf{r}) + \ldots + \mathbf{E}_n(\mathbf{r})$

Writing out each term:

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{q_1}{r_{1P}^2} \hat{\mathbf{r}}_{1P} + \frac{1}{4\pi\varepsilon_0} \frac{q_2}{r_{2P}^2} \hat{\mathbf{r}}_{2P} + \ldots + \frac{1}{4\pi\varepsilon_0} \frac{q_n}{r_{nP}^2} \hat{\mathbf{r}}_{nP}$

Since the factor $\frac{1}{4\pi\varepsilon_0}$ is common to every term, pull it out front. This gives the compact summation:

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{\mathbf{r}}_{iP} \qquad \text{(1.10)}$

Reading the Formula: What Each Symbol Means

Symbol	Meaning
$\mathbf{E}(\mathbf{r})$	Total electric field at the point P located at position $\mathbf{r}$
$q_i$	The $i$-th source charge (can be positive or negative)
$r_{iP}$	Distance from charge $q_i$ to the point P
$\hat{\mathbf{r}}_{iP}$	Unit vector pointing from $q_i$ toward P
$\varepsilon_0$	Permittivity of free space ($8.854 \times 10^{-12}$ C$^2$ N$^{-1}$ m$^{-2}$)

Key Features of the Result

• $\mathbf{E}$ is a vector quantity. It has both magnitude and direction at every point in space, and both can change from one location to another.
• It depends only on the source charges. The positions and magnitudes of $q_1, q_2, \ldots, q_n$ fully determine the field everywhere. No test charge is involved in the formula.
• The sign of each $q_i$ automatically handles direction. A positive charge pushes the field outward (away from itself), while a negative charge pulls it inward (toward itself). The vector sum takes care of all cancellations and reinforcements.
• This is a vector field. In physics, the word “field” refers to a quantity defined at every point in space. Since the electric field assigns a vector to each point, it is called a vector field.

Why the Electric Field Is More Than a Mathematical Shortcut

You might wonder: why introduce the electric field at all? If you know where every charge sits, you can compute the force on any charge directly using Coulomb’s law and superposition. Why add this extra layer?

In Electrostatics: Convenient but Not Essential

For stationary charges, the field concept is indeed a convenience rather than a necessity. It gives you an elegant way to describe the electrical environment: the field at any point tells you the force a unit positive test charge would feel if placed there. It captures the combined effect of all source charges in one location-dependent vector, independent of any test charge you might use.

But if convenience were its only virtue, the field would be just a mathematical shorthand, not a physical entity.

In Time-Dependent Situations: Physically Indispensable

The real significance of the electric field shows up when charges start to move, especially when they accelerate.

Picture two charges $q_1$ and $q_2$ far apart. Now let $q_1$ begin accelerating. Does $q_2$ instantly feel a different force? It cannot, because the fastest any signal or influence can travel is $c$, the speed of light. There must be a time lag between the cause (the motion of $q_1$) and the effect (the changed force on $q_2$).

The field picture handles this naturally:

1. When $q_1$ accelerates, it disturbs the electromagnetic field in its vicinity.
2. That disturbance propagates outward as an electromagnetic wave, travelling at speed $c$.
3. When the wave reaches $q_2$, it exerts a force on $q_2$.

No mysterious instant “action at a distance” is needed. The field itself carries the information, and the finite travel time of the wave explains the delay.

Fields as Real Physical Entities

This picture elevates electric and magnetic fields from abstract tools to genuine physical entities. Three properties cement their reality:

• Independent dynamics — fields evolve according to their own laws (later formalised as Maxwell’s equations), not simply as passive reflections of where charges happen to be.
• Energy transport — electromagnetic fields can carry energy through space. A source of time-varying fields that is switched on briefly and then turned off leaves behind propagating waves that continue to carry energy outward, even though the source is no longer active.
• Detection through effects — we detect fields only via the forces they exert on charges, but this does not make them any less real. They exist, evolve, and transport energy on their own.

The concept of the field was first introduced by Faraday and has since become one of the most central ideas in all of physics.

Worked Example 1.7: Electron and Proton Falling in a Uniform Field

Problem: An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude $2.0 \times 10^4$ N/C. The field direction is then reversed (keeping the same magnitude), and a proton falls through the same distance. Find the time of fall for each particle, and compare with free fall under gravity.

Fig 1.10: (a) Electron falling in a uniform electric field (b) Proton falling in a reversed uniform electric field

The Electron

The electric field points upward. Because the electron carries a negative charge, the electric force on it acts opposite to the field, i.e. downward. The magnitude of this force is:

$F_e = eE$

where $e = 1.6 \times 10^{-19}$ C is the magnitude of the electron’s charge and $E = 2.0 \times 10^4$ N/C is the field strength.

Applying Newton’s second law ($F = ma$), the acceleration of the electron is:

$a_e = \frac{eE}{m_e}$

where $m_e = 9.11 \times 10^{-31}$ kg.

The electron starts from rest and falls through a distance $h = 1.5 \times 10^{-2}$ m. Using the kinematic equation for motion from rest ($h = \frac{1}{2}at^2$):

$h = \frac{1}{2} a_e \, t_e^2$

Rearranging for $t_e$:

$t_e^2 = \frac{2h}{a_e}$

$t_e = \sqrt{\frac{2h}{a_e}}$

Substituting $a_e = eE/m_e$:

$t_e = \sqrt{\frac{2h \, m_e}{eE}}$

Plugging in values:

$t_e = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}}$

Numerator: $2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31} = 2.733 \times 10^{-32}$

Denominator: $1.6 \times 10^{-19} \times 2.0 \times 10^4 = 3.2 \times 10^{-15}$

$t_e = \sqrt{\frac{2.733 \times 10^{-32}}{3.2 \times 10^{-15}}} = \sqrt{8.54 \times 10^{-18}}$

$\boxed{t_e \approx 2.9 \times 10^{-9} \text{ s}}$

The Proton

Now the field is reversed to point downward. The proton carries a positive charge, so the electric force on it acts in the same direction as the field: downward. The force magnitude is still $eE$ (same charge magnitude), but the proton’s mass is much larger.

The acceleration of the proton is:

$a_p = \frac{eE}{m_p}$

where $m_p = 1.67 \times 10^{-27}$ kg.

Using the same kinematic equation:

$t_p = \sqrt{\frac{2h \, m_p}{eE}}$

$t_p = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}}$

Numerator: $2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27} = 5.01 \times 10^{-29}$

Denominator: $3.2 \times 10^{-15}$ (same as before)

$t_p = \sqrt{\frac{5.01 \times 10^{-29}}{3.2 \times 10^{-15}}} = \sqrt{1.566 \times 10^{-14}}$

$\boxed{t_p \approx 1.3 \times 10^{-7} \text{ s}}$

Why This Differs Fundamentally From Free Fall Under Gravity

Under gravity, the time to fall a certain distance does not depend on the mass of the falling object. A feather and a cannonball (in vacuum) hit the ground at the same time. This happens because the gravitational force is proportional to mass ($F = mg$), so the acceleration $a = F/m = g$ is the same for all objects.

In an electric field, the situation changes completely. The electric force depends on charge, not mass: $F = eE$. Since the electron and proton carry the same magnitude of charge, they feel the same force. But their masses differ by a factor of about 1836. Acceleration $a = eE/m$ is therefore very different for the two, and the heavier proton takes far longer.

Checking Whether Gravity Can Be Ignored

The electric acceleration of the proton is:

$a_p = \frac{eE}{m_p} = \frac{(1.6 \times 10^{-19})(2.0 \times 10^4)}{1.67 \times 10^{-27}}$

$a_p = \frac{3.2 \times 10^{-15}}{1.67 \times 10^{-27}} \approx 1.9 \times 10^{12} \text{ m/s}^2$

This is roughly $2 \times 10^{11}$ times larger than $g = 9.8$ m/s$^2$. The electron’s electric acceleration is even larger. So gravitational effects are entirely negligible in this problem.

Worked Example 1.8: Electric Fields at Three Points Around Two Charges

Problem: Two point charges, $q_1 = +10^{-8}$ C (positive) and $q_2 = -10^{-8}$ C (negative), are placed 0.1 m apart on a horizontal line, with $q_1$ on the left and $q_2$ on the right. Calculate the electric field at three points:

• Point A: the midpoint between the two charges (0.05 m from each)
• Point B: located 0.05 m to the left of $q_1$ (so 0.15 m from $q_2$)
• Point C: at the apex of an equilateral triangle with $q_1$ and $q_2$ as the base (0.1 m from each charge)

Fig 1.11: Two point charges $q_1$ (positive) and $q_2$ (negative) separated by 0.1 m, with field calculation points A, B, and C

At Point A (the Midpoint)

Point A is equidistant from both charges, 0.05 m from each.

Field from $q_1$ (positive): The field from a positive charge points away from it. Since A is to the right of $q_1$, this field points to the right.

$E_{1A} = \frac{k|q_1|}{r_{1A}^2} = \frac{(9 \times 10^9)(10^{-8})}{(0.05)^2} = \frac{90}{0.0025} = 3.6 \times 10^4 \text{ N/C}$

Field from $q_2$ (negative): The field from a negative charge points toward it. Since $q_2$ is to the right of A, this field also points to the right.

The magnitude is the same (equal charge magnitude, same distance):

$E_{2A} = 3.6 \times 10^4 \text{ N/C}$

Both fields point in the same direction (to the right), so they reinforce each other:

$E_A = E_{1A} + E_{2A} = 3.6 \times 10^4 + 3.6 \times 10^4 = 7.2 \times 10^4 \text{ N/C}$

$\boxed{\mathbf{E}_A = 7.2 \times 10^4 \text{ N/C, directed to the right}}$

Notice that between a positive and a negative charge, the fields from both charges point in the same direction (from the positive toward the negative). They add up rather than cancel.

At Point B (to the Left of $q_1$)

Point B is 0.05 m from $q_1$ (on the far side, away from $q_2$) and therefore 0.15 m from $q_2$.

Field from $q_1$ (positive): Points away from $q_1$. Since B is to the left of $q_1$, this field points to the left.

$E_{1B} = \frac{(9 \times 10^9)(10^{-8})}{(0.05)^2} = 3.6 \times 10^4 \text{ N/C}$

Field from $q_2$ (negative): Points toward $q_2$. Since $q_2$ is to the right of B, this field points to the right.

$E_{2B} = \frac{(9 \times 10^9)(10^{-8})}{(0.15)^2} = \frac{90}{0.0225} = 4 \times 10^3 \text{ N/C}$

The two fields point in opposite directions. The leftward field ($E_{1B}$) is much stronger because B is closer to $q_1$ than to $q_2$:

$E_B = E_{1B} - E_{2B} = 3.6 \times 10^4 - 4 \times 10^3 = 3.2 \times 10^4 \text{ N/C}$

$\boxed{\mathbf{E}_B = 3.2 \times 10^4 \text{ N/C, directed to the left}}$

At Point C (Equilateral Triangle Apex)

Point C is 0.1 m from both charges (the side length of the equilateral triangle equals the charge separation). Each interior angle of the equilateral triangle is $60°$.

Magnitude of each individual field at C:

$E_{1C} = E_{2C} = \frac{(9 \times 10^9)(10^{-8})}{(0.10)^2} = \frac{90}{0.01} = 9 \times 10^3 \text{ N/C}$

The two field vectors at C point in different directions:

• $\mathbf{E}_{1C}$ (from positive $q_1$) points away from $q_1$, which is upward and to the right.
• $\mathbf{E}_{2C}$ (from negative $q_2$) points toward $q_2$, which is downward and to the right.

To find the resultant, resolve each vector along the horizontal direction (the line joining the charges). Each field vector makes an angle of $60°$ with the horizontal. The horizontal (rightward) components add:

$E_{C,\text{horizontal}} = E_{1C} \cos 60° + E_{2C} \cos 60°$

$= 9 \times 10^3 \times \frac{1}{2} + 9 \times 10^3 \times \frac{1}{2} = 9 \times 10^3 \text{ N/C}$

The vertical components point in opposite directions (one upward, one downward) with equal magnitudes, so they cancel exactly.

$\boxed{\mathbf{E}_C = 9 \times 10^3 \text{ N/C, directed to the right}}$

Patterns to Notice

This example illustrates how the direction and arrangement of charges lead to very different field behaviours at different points:

• At the midpoint between opposite charges: the fields reinforce (both point from $+$ toward $-$), giving a large resultant.
• On the far side of one charge: the nearer charge dominates, but the farther charge partially cancels.
• At the equilateral apex: symmetry causes vertical components to cancel, leaving only a horizontal resultant.