Coulomb Law and Electric Field
Coulomb's Law and Electrostatic Force
Coulomb's Law quantifies the mutual electrostatic force between two stationary point charges as proportional to the product of charges and inversely proportional to the square of their separation.
$$F = k \frac{|q_1 q_2|}{r^2}$$
Determines the magnitude of force between two static charges.
$F$=Electrostatic force(N)
$k$=Coulomb's constant = $\frac{1}{4\pi\epsilon_0}$ ≈ 8.99 × 10⁹(N·m²/C²)
$q_1, q_2$=Charge magnitudes(C)
$r$=Distance between centers of the point charges(m)
$r \to 0$
→Force approaches infinity (point charge idealization breaks down)
$q_1, q_2$ same sign
→Repulsive force (positive product)
$q_1, q_2$ opposite sign
→Attractive force (negative product)
Inverse Square Scaling: Doubling $r$ reduces force to $1/4$; tripling $r$ reduces it to $1/9$. The pattern is $F' = F/n^2$ when distance is multiplied by $n$.
Permittivity of Free Space: $\epsilon_0 = 8.85 \times 10^{-12}$ C²/(N·m²) gives $k = 1/(4\pi\epsilon_0) = 9 \times 10^9$ N·m²/C².
Newton's Third Law: Coulomb's force is mutual — $\vec{F}_{12} = -\vec{F}_{21}$. Both charges experience equal and opposite forces regardless of their magnitudes.
Dimensional Check: $[k][q^2]/[r^2]$ = (N·m²/C²)(C²)/(m²) = N. Always verify your answer has units of force.
The vector form of Coulomb's Law gives both the magnitude and direction of the force using unit vectors along the line joining the charges.
$$\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{21}$$
Gives the force on charge 2 due to charge 1, including direction via the unit vector.
$\vec{F}_{21}$=Force on $q_2$ due to $q_1$(N)
$\hat{r}_{21}$=Unit vector from $q_1$ toward $q_2$(dimensionless)
Both charges positive
→Force is along $\hat{r}_{21}$ (repulsion, away from source)
Opposite charges
→Product $q_1 q_2 < 0$ reverses direction (attraction, toward source)
Unit Vector Direction: $\hat{r}_{21}$ points from the source charge ($q_1$) to where the force is felt ($q_2$). Swapping subscripts reverses direction: $\hat{r}_{12} = -\hat{r}_{21}$.
Sign Convention: In the vector form, keep the signs of $q_1$ and $q_2$. The formula automatically gives the correct direction — positive product means repulsion, negative means attraction.
When a dielectric material (insulator) fills the space between charges, the electrostatic force is reduced by the material's relative permittivity $\epsilon_r$.
$$F = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$$
Coulomb's force in a medium is weaker than in vacuum by a factor of $\epsilon_r$.
$\epsilon_r$=Relative permittivity (dielectric constant) of the medium(dimensionless)
Vacuum ($\epsilon_r = 1$)
→Reduces to the standard Coulomb's law
Water ($\epsilon_r \approx 78.5$)
→Force is reduced to about $1/78$ of vacuum value
Air ($\epsilon_r \approx 1.0006$)
→Essentially the same as vacuum — safely use $\epsilon_r = 1$
Force Ratio Shortcut: $F_{\text{medium}} = F_{\text{vacuum}} / \epsilon_r$. No need to recalculate from scratch — just divide.
Why It Decreases: The dielectric's molecules partially align with the external field, creating an opposing internal field that effectively shields the charges from each other.
Key Values: Vacuum = 1, Air ≈ 1, Glass ≈ 5–10, Water ≈ 78.5. Water's high $\epsilon_r$ is why ionic compounds dissolve easily in it.
The Concept of the Electric Field
The electric field intensity (E) is defined as the force per unit positive test charge placed at a point, representing the property of space established by a source charge.
$$\vec{E} = \frac{\vec{F}}{q_0}$$
Defines field strength as a vector quantity independent of the test charge magnitude.
$\vec{E}$=Electric field vector(N/C (equivalently V/m))
$\vec{F}$=Force on test charge(N)
$q_0$=Magnitude of the positive test charge(C)
Positive charge in field
→Force is parallel to $\vec{E}$ ($\vec{F} = q\vec{E}$)
Negative charge in field
→Force is antiparallel to $\vec{E}$ ($\vec{F} = q\vec{E}$, but $q < 0$)
Field Exists Independently: The source charge creates a field whether or not another charge is there to feel it. The test charge merely probes what already exists.
Test Charge Must Be Small: If $q_0$ is too large, its own field rearranges the source charges, changing the very field you're trying to measure.
Dimensional Check: $[F]/[q]$ = N/C. The field can also be expressed as V/m (equivalent unit).
Electric Field of a Point Charge
A single source charge $Q$ creates a field that spreads radially with a magnitude determined by the distance $r$ from the charge center.
$$E = k \frac{|Q|}{r^2}$$
Calculates the field strength generated by a single point charge at distance $r$.
$E$=Electric field strength(N/C)
$Q$=Source charge(C)
$r$=Distance from source charge(m)
$Q > 0$
→Field points radially outward from the charge
$Q < 0$
→Field points radially inward toward the charge
$r \to \infty$
→Field approaches zero — charge influence fades with distance
Radial Symmetry: At a fixed distance $r$, the magnitude of $E$ is the same in every direction. Only the direction (radially outward or inward) changes with the sign of $Q$.
Proportionality Shortcut: If you know $E_1$ at distance $r_1$, then at distance $r_2$: $E_2 = E_1 (r_1/r_2)^2$. No need to recalculate from scratch.
Combining Force and Field: Once you know $\vec{E}$ at a point, the force on any charge $q$ placed there is simply $\vec{F} = q\vec{E}$.
Superposition Principle
The total electric field at any point is the vector sum of individual fields produced by each charge independently — this is the superposition principle.
$$\vec{E}_{net} = \sum_i \vec{E}_i$$
Total field is the cumulative vector sum of all contributing source fields.
$\vec{E}_{net}$=Net electric field vector at the point(N/C)
$\vec{E}_i$=Field contribution from the $i$-th charge(N/C)
Two equal like charges, midpoint
→Fields cancel → $\vec{E}_{net} = 0$ at midpoint
Two equal unlike charges, midpoint
→Fields add (both point toward the negative charge)
Vector Addition Required: You cannot just add magnitudes. Break each $\vec{E}_i$ into components, sum $x$-components and $y$-components separately, then recombine.
Zero-Field Point (Two Like Charges): Between two unequal positive charges $q_1 > q_2$, the null point is closer to the smaller charge. Set $E_1 = E_2$ and solve for position.
Zero-Field Point (Two Unlike Charges): The null point lies outside the pair, on the side of the smaller magnitude charge.
Electric Field Lines
Electric field lines are a visual map of the field: they show the direction of $\vec{E}$ at every point and their spacing indicates the field strength.
Origin and Termination: Field lines start on positive charges and end on negative charges. An isolated positive charge has lines going to infinity.
Tangent Rule: The tangent to a field line at any point gives the direction of $\vec{E}$ at that point.
Density ∝ Strength: Where lines are closely packed, the field is strong; where they spread apart, the field is weak.
Non-Crossing Rule: Field lines never intersect. If they did, $\vec{E}$ would have two directions at the crossing point, which is physically impossible.
Uniform Field: Parallel, equally spaced lines (e.g., between parallel plates of a capacitor) represent a uniform electric field — same $\vec{E}$ everywhere.
Common field line patterns include the radial field of a single charge, the dipole pattern of two opposite charges, and the repulsion pattern of two like charges.
Single Positive Charge: Lines radiate outward uniformly in all directions.
Single Negative Charge: Lines point radially inward from all directions.
Dipole (±q): Lines curve from the positive charge to the negative charge. The field is strongest along the axis connecting them.
Two Like Charges: Lines repel in the middle region. A neutral zone (zero-field point) exists between them.
Parallel Plates: Uniform, parallel lines in the central region; fringe effects at the edges.
Applications of Electrostatics
A photocopier (xerography) uses electrostatic principles: a selenium-coated drum retains charge in dark areas (image pattern), attracts oppositely charged toner, and transfers it to paper.
Photoconductor: Selenium is an insulator in the dark but becomes a conductor when exposed to light. Light areas lose their charge; dark areas retain it.
Toner Transfer: Negatively charged toner sticks to positively charged (dark) areas on the drum, then transfers to paper and is fused by heated rollers.
Inkjet printers use a charging electrode and deflection plates (a parallel-plate capacitor) to steer ink droplets: uncharged drops hit the paper, while charged drops are deflected into a gutter.
Deflection Mechanism: Charged droplets experience a force $\vec{F} = q\vec{E}$ between the deflection plates, curving them away from the paper into a collection gutter.
Print Control: The computer turns the charging electrode on/off — uncharged droplets fly straight to paper, charged ones are diverted.
Application Strategy
Worked Examples
Problem Statement
Calculate the magnitude of the electrostatic force between two protons separated by $5.0 \times 10^{-11}$ m.
Coulomb's LawElementary Charge
Strategic Toolkit
$$F = k \frac{q_1 q_2}{r^2}$$
Step 1
Identify Constants
The charge of a proton is $e = 1.6 \times 10^{-19}$ C and $k = 8.99 \times 10^9$ N·m²/C².
$q_1 = q_2 = 1.6 \times 10^{-19}$ C
Step 2
Substitution
Plug the values into Coulomb's equation.
$$F = \frac{(8.99 \times 10^9)(1.6 \times 10^{-19})^2}{(5.0 \times 10^{-11})^2}$$
Step 3
Calculate Magnitude
Evaluate: numerator = $8.99 \times 10^9 \times 2.56 \times 10^{-38} = 2.30 \times 10^{-28}$, denominator = $2.5 \times 10^{-21}$.
$F \approx 9.2 \times 10^{-8}$ N
Problem Statement
Find the net electric field at point P, $0.3$ m right of a $+2.0$ μC charge and $0.2$ m left of a $-5.0$ μC charge.
Point Charge E-FieldVector Superposition
Strategic Toolkit
$$E = k\frac{|Q|}{r^2}$$
Step 1
Field from Q1
Calculate field from $+2$ μC. Direction is Right (away from positive).
$E_1 = \frac{(9 \times 10^9)(2 \times 10^{-6})}{(0.3)^2} = 2.0 \times 10^5$ N/C
Step 2
Field from Q2
Calculate field from $-5$ μC. Direction is Right (toward negative charge).
$E_2 = \frac{(9 \times 10^9)(5 \times 10^{-6})}{(0.2)^2} = 1.125 \times 10^6$ N/C
Step 3
Sum Vectors
Both fields point Right at P (away from +Q, toward −Q), so add magnitudes.
$E_{net} = 2.0 \times 10^5 + 1.125 \times 10^6 = 1.33 \times 10^6$ N/C (Right)
Problem Statement
Two positive charges $q_1 = 16$ μC and $q_2 = 4$ μC are 3.0 m apart. Find the point on the line between them where the electric field is zero.
SuperpositionZero-Field Condition
Strategic Toolkit
$$\frac{kq_1}{(3-d)^2} = \frac{kq_2}{d^2}$$
Step 1
Set Up Equation
Let $d$ be the distance from $q_2$. At the null point, field magnitudes are equal but directions opposite.
$$\frac{q_1}{(3-d)^2} = \frac{q_2}{d^2}$$
Step 2
Substitute and Simplify
Plug in charge values and cross-multiply.
$16d^2 = 4(3-d)^2 \Rightarrow 4d^2 = (3-d)^2 \Rightarrow 2d = 3-d$
Step 3
Solve
Solve $2d = 3 - d$ → $3d = 3$ → $d = 1.0$ m from $q_2$ (the smaller charge). The null point is always closer to the weaker charge.
$d = 1.0$ m from $q_2$
Problem Statement
Two charges of $+3$ μC and $-3$ μC are 10 cm apart in water ($\epsilon_r = 78.5$). What is the force between them?
Coulomb's LawRelative Permittivity
Strategic Toolkit
$$F = \frac{1}{4\pi\epsilon_0 \epsilon_r}\frac{q_1 q_2}{r^2}$$
Step 1
Force in Vacuum
First calculate what the force would be in vacuum.
$F_0 = \frac{(9 \times 10^9)(3 \times 10^{-6})^2}{(0.1)^2} = 8.1$ N
Step 2
Apply Dielectric Factor
Divide by $\epsilon_r$ to get force in water.
$F = F_0 / \epsilon_r = 8.1 / 78.5 = 0.103$ N
Step 3
Result
The force is attractive (opposite charges) and roughly 78 times weaker than in vacuum.
$F \approx 0.10$ N (attractive)
Memorize formulas
Coulomb's Law
Inverse square interaction between point charges in vacuum.
Formula
$$F = k \frac{|q_1 q_2|}{r^2}$$
Coulomb's Law in Medium
Force reduced by the dielectric constant of the medium.
Formula
$$F = \frac{1}{4\pi\epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$$
Electric Field Definition
Force per unit positive test charge.
Formula
$$\vec{E} = \frac{\vec{F}}{q_0}$$
Point Charge E-Field
Field magnitude at distance $r$ from a source charge.
Formula
$$E = k \frac{|Q|}{r^2}$$
Review
1
Two point charges are separated by a distance $d$. If one charge is doubled and the distance is tripled, by what factor does the electrostatic force change?
Ans
Decreases to $2/9$ of original
Review
Tap to reveal
Apply Coulomb's law scaling
The force is proportional to the product of the charges and inversely proportional to the square of the separation. So doubling one charge multiplies the force by 2, while tripling the distance multiplies it by $1/3^2$.
$$\frac{F'}{F} = 2 \times \frac{1}{3^2} = \frac{2}{9}$$
2
Charges $+4$ μC and $-4$ μC are placed 0.2 m apart. What is the magnitude of the net electric field at the midpoint between them?
Ans
$7.2 \times 10^6$ N/C
Review
Tap to reveal
Find the field from one charge at the midpoint
The midpoint is 0.1 m from each charge, so use $E = k|q|/r^2$ for either charge.
$$E_1 = \frac{(9 \times 10^9)(4 \times 10^{-6})}{(0.1)^2} = 3.6 \times 10^6\ \text{N/C}$$
Add the fields because they point in the same direction
At the midpoint, the field from the positive charge points away from it and the field from the negative charge points toward it. Both directions are from $+$ toward $-$, so the magnitudes add.
$$E_{net} = 2E_1 = 2(3.6 \times 10^6) = 7.2 \times 10^6\ \text{N/C}$$
3
A positive test charge is placed in an electric field. If the magnitude of the test charge is doubled, the force on it doubles. What happens to the measured electric field strength $E$ at that location?
Ans
It remains the same
Review
Tap to reveal
Use the definition of electric field
Electric field strength is defined by $E = F/q_0$. If the test charge is doubled, the force also doubles, so the ratio stays unchanged.
$$E' = \frac{2F}{2q_0} = \frac{F}{q_0} = E$$
4
Two charges experience a force of 5 N when separated by a dielectric medium with $\epsilon_r = 6$. What would the force be if the dielectric is removed and they are placed in vacuum at the same distance?
Ans
30 N
Review
Tap to reveal
Reverse the dielectric reduction
A dielectric reduces the electrostatic force by a factor of $\epsilon_r$. To recover the vacuum force, multiply the measured force by $\epsilon_r$.
$$F_{vacuum} = F_{medium}\,\epsilon_r = 5 \times 6 = 30\ \text{N}$$
5
A point charge $Q$ produces a field of $E_0$ at distance $r$. At what distance is the field $E_0/4$?
Ans
$2r$
Review
Tap to reveal
Use the inverse-square dependence
For a point charge, $E \propto 1/r^2$. To make the field one-fourth as large, the distance must increase so that $r^2$ becomes 4 times larger.
$$\frac{E_0/4}{E_0} = \frac{r^2}{r'^2} = \frac{1}{4} \Rightarrow r' = 2r$$
6
An electron is placed in a uniform electric field pointing East with magnitude $1000$ N/C. What is the magnitude and direction of the force on the electron?
Ans
$1.6 \times 10^{-16}$ N, West
Review
Tap to reveal
Calculate the force magnitude
Use $F = |q|E$ with the electron charge magnitude and the given field strength.
$$F = (1.6 \times 10^{-19})(1000) = 1.6 \times 10^{-16}\ \text{N}$$
Reverse the direction for a negative charge
A negative charge experiences force opposite to the electric field direction. Since the field points East, the force points West.
$$1.6 \times 10^{-16}\ \text{N, West}$$
7
The electrostatic force between two charges is $F$ when they are in vacuum. If the same charges are placed in water ($\epsilon_r = 78.5$) and the distance is halved, the new force is approximately:
Ans
$F/19.6$
Review
Tap to reveal
Combine the distance and medium factors
Halving the separation increases the force by a factor of 4. Water reduces the force by a factor of $78.5$, so multiply the two effects.
$$F' = F \times \frac{4}{78.5} \approx \frac{F}{19.6}$$
8
Two identical positive charges $Q$ are at opposite corners of a square of side $a$. What is the magnitude of the electric field at the center of the square?
Ans
Zero
Review
Tap to reveal
Use symmetry at the square's center
The center is the same distance from both identical charges, so each produces an equal-magnitude field there. The directions are opposite along the diagonal, so the vectors cancel exactly.
$$E_{net} = 0$$
9
Charge $q_1 = +5$ μC is at the origin. What is the electric field magnitude at a point 0.5 m away?
Ans
$1.8 \times 10^5$ N/C
Review
Tap to reveal
Substitute into the point-charge field formula
For a point charge, the field magnitude is $E = kQ/r^2$. Substitute the given values and square the distance correctly.
$$E = \frac{(9 \times 10^9)(5 \times 10^{-6})}{(0.5)^2} = \frac{4.5 \times 10^4}{0.25} = 1.8 \times 10^5\ \text{N/C}$$
10
In a photocopier, the drum is coated with selenium because selenium is:
Ans
A photoconductor — insulator in the dark, conductor in light
Review
Tap to reveal
Identify the required material property
A photocopier needs a surface that can hold charge in the dark and lose charge when light falls on it. That behavior is photoconductivity.
Photoconductor: insulator in the dark, conductor in light
11
Two charges of $+3$ μC and $-3$ μC are placed 0.1 m apart. At the midpoint between them, the electric field is:
Ans
$1.08 \times 10^7$ N/C, directed from $+$ toward $-$
Review
Tap to reveal
Compute the field from one charge at the midpoint
The midpoint lies 0.05 m from each charge. Use $E = k|q|/r^2$ for one charge first.
$$E_1 = \frac{(9 \times 10^9)(3 \times 10^{-6})}{(0.05)^2} = 1.08 \times 10^7\ \text{N/C}$$
Add the two fields in the same direction
For opposite charges, both field vectors at the midpoint point from the positive charge toward the negative charge, so they reinforce rather than cancel.
$$E_{net} = 2E_1 = 1.08 \times 10^7\ \text{N/C, from } + \text{ toward } -$$
Question 1 of 11
0:00
1
Two point charges are separated by a distance $d$. If one charge is doubled and the distance is tripled, by what factor does the electrostatic force change?