Electric Field Formula: 9 Equations, Units, and Worked Examples

The electric field formula is E = F/q. It states that electric field strength equals the force on a test charge divided by the charge. The 9 formulas below cover point charges, parallel plates, lines, sheets, spheres, rings, and dipoles.

For a reliable reference, see OpenStax defines electric field as force per unit charge and explains fields from point charges.

The electric field also describes force per unit charge; see Electric Force Equation: Coulomb’s Law, 7 Formulas, and Examples.

What Is the Electric Field Formula?

The electric field formula is E = F/q, where E is electric field strength, F is the electric force on a test charge, and q is the test charge.

Electric field strength is a vector quantity. It is also called electric field intensity. It measures the force per unit charge at a point in space. A test charge is a small positive charge that does not disturb the source charges. According to OpenStax University Physics Volume 2, the electric field is defined through the force on such a test charge.

Sources of electric fields include electrons, protons, charged plates, and charged spheres. The field exists at every point around a source charge, whether or not a test charge is present.

What Do the Symbols in the Electric Field Formula Represent?

The electric field equations use 15 symbols, and each symbol has a defined SI unit. Table 1 lists the symbols, quantity names, SI units, and values.

Table 1: Symbols in electric field equations

The NIST CODATA 2022 recommended values list ε₀ as 8.8541878188 × 10⁻¹² F/m. The 2019 revision of the SI fixed the elementary charge at exactly 1.602176634 × 10⁻¹⁹ C. As a result, ε₀ is measured experimentally and is no longer an exact value.

What Are the 9 Main Electric Field Formulas?

The 9 main electric field formulas apply to a test charge, a point charge, parallel plates, a line, a sheet, a sphere, a ring, and a dipole. Table 2 lists each formula with its field source and valid condition.

Table 2: Electric field formulas by field source

What Is the Electric Field Formula From Force?

E = F/q defines electric field as force per unit positive test charge. The formal definition is the limit of F/q as q approaches zero, which keeps the test charge from disturbing the source charges. The unit is N/C. The field direction matches the force direction on a positive test charge.

What Is the Electric Field Formula for a Point Charge?

E = kQ/r² gives the field magnitude at distance r from a point charge Q. The formula follows from Coulomb’s law, F = kQq/r², after dividing the force by q. Charles-Augustin de Coulomb published the inverse-square force law in 1785, based on torsion balance measurements. Doubling r reduces E to one fourth. Tripling r reduces E to one ninth.

What Is the Electric Field Formula Between Two Parallel Plates?

E = V/d gives the uniform field between parallel plates, where V is the potential difference and d is the plate separation. The charge form is E = σ/ε₀. Both forms agree, because V = σd/ε₀ for ideal plates. The field points from the positive plate to the negative plate. Edge effects are ignored, if d is small compared with the plate dimensions.

What Is the Electric Field Formula for a Line of Charge?

E = λ/(2πε₀r) gives the field at perpendicular distance r from an infinite line with linear charge density λ. The equivalent form is E = 2kλ/r. The field decreases in proportion to 1/r. Gauss’s law with a cylindrical surface produces this result.

What Is the Electric Field Formula for a Sheet of Charge?

E = σ/(2ε₀) gives the field of an infinite non-conducting sheet with uniform surface charge density σ. The field strength is independent of the distance from the sheet. Two oppositely charged sheets produce E = σ/ε₀ between them and zero outside.

What Is the Electric Field Formula for a Charged Sphere?

Outside a charged sphere, E = kQ/r². Inside a conducting sphere in equilibrium, E = 0. A uniformly charged solid insulating sphere of radius R has E = kQr/R³ for r < R. The field reaches its maximum value, kQ/R², at the surface. Gauss’s law derives these results.

What Is the Electric Field Formula for a Ring of Charge?

E = kQx/(x² + R²)^(3/2) gives the field on the axis of a uniformly charged ring of radius R at distance x from its center. The field is zero at the ring center, because opposite sides cancel. For x much larger than R, the formula reduces to E = kQ/x².

What Is the Electric Field Formula for an Electric Dipole?

E ≈ 2kp/r³ gives the field on the dipole axis, and E ≈ kp/r³ gives the field on the perpendicular bisector, for r much larger than the charge separation a. The dipole moment is p = Qa. Dipole fields decrease in proportion to 1/r³, which is faster than the 1/r² decrease of a point charge. Water molecules are examples of dipoles, with a dipole moment of about 6.2 × 10⁻³⁰ C·m.

What Is the General Electric Field Formula for Any Charge Distribution?

The general electric field formula is the vector sum E = E₁ + E₂ + … + Eₙ, where each term is the field of one charge. For a continuous distribution, the sum becomes the integral E = k ∫ (dq/r²) r̂, where dq is a small charge element and r̂ is the unit vector from the element to the field point.

What Is the Superposition Principle for Electric Fields?

The superposition principle states that the net electric field at a point equals the vector sum of the fields from every individual charge. The components add separately: Ex = ΣEix and Ey = ΣEiy. The net magnitude is E = √(Ex² + Ey²).

What Is Gauss’s Law for Electric Fields?

Gauss’s law is ∮ E · dA = Qenc/ε₀, where Qenc is the charge enclosed by a closed surface. The law is one of the 4 Maxwell equations. It simplifies field calculations for spherical, cylindrical, and planar symmetry.

How Is the Electric Field Related to Electric Potential?

E = −∇V relates the electric field to the electric potential. In one dimension, E = −dV/dx. For a uniform field, the magnitude reduces to E = V/d. The field points in the direction of decreasing potential.

What Is the Direction of an Electric Field?

An electric field points away from positive charges and toward negative charges. The direction at any point equals the direction of the force on a positive test charge. A negative charge experiences a force opposite to the field. Michael Faraday introduced field lines in the 19th century to represent this direction.

Electric field lines follow 4 rules:

• Field lines start on positive charges.
• Field lines end on negative charges.
• Field lines never cross.
• Field lines are closer together where the field is stronger.

What Are the Units of the Electric Field?

The SI unit of electric field is the newton per coulomb (N/C), which equals the volt per meter (V/m). The equivalence follows from 1 V = 1 J/C and 1 J = 1 N·m. In SI base units, the electric field unit is kg·m·s⁻³·A⁻¹.

How Do You Calculate an Electric Field Step by Step?

To calculate an electric field, identify the source type, select the matching formula, convert the units, substitute the values, and assign the direction. The 6 steps are:

1. Identify the source type: point charge, plates, line, sheet, sphere, ring, or dipole.
2. Select the matching formula from Table 2.
3. Convert all values to SI units: meters, coulombs, and volts.
4. Substitute the values and compute the magnitude.
5. Assign the direction: away from positive charges and toward negative charges.
6. Add the fields as vectors, if several charges contribute.

What Are 4 Worked Examples of the Electric Field Formula?

The 4 worked examples below apply the electric field formula to a point charge, a measured force, parallel plates, and two opposite charges. All examples use k = 8.99 × 10⁹ N·m²/C².

How Do You Find the Electric Field of a 2.0 μC Point Charge at 0.50 m?

• Given: Q = 2.0 × 10⁻⁶ C and r = 0.50 m.
• Formula: E = kQ/r².
• Calculation: E = (8.99 × 10⁹ × 2.0 × 10⁻⁶) / (0.50)² = 17,980 / 0.25.
• Answer: E = 7.2 × 10⁴ N/C, directed radially outward.

How Do You Find the Electric Field From a Measured Force?

• Given: F = 3.0 × 10⁻³ N and q = 1.5 × 10⁻⁶ C.
• Formula: E = F/q.
• Calculation: E = (3.0 × 10⁻³) / (1.5 × 10⁻⁶).
• Answer: E = 2.0 × 10³ N/C, directed along the force on the positive test charge.

How Do You Find the Electric Field Between Parallel Plates With 12 V Across 4.0 mm?

• Given: V = 12 V and d = 4.0 mm = 4.0 × 10⁻³ m.
• Formula: E = V/d.
• Calculation: E = 12 / (4.0 × 10⁻³).
• Answer: E = 3.0 × 10³ V/m, directed from the positive plate to the negative plate.
• Force on an electron: F = eE = (1.602 × 10⁻¹⁹ C)(3.0 × 10³ N/C) = 4.8 × 10⁻¹⁶ N, directed toward the positive plate.

How Do You Find the Electric Field at the Midpoint Between +1.0 μC and −1.0 μC Charges?

• Given: Q₁ = +1.0 × 10⁻⁶ C, Q₂ = −1.0 × 10⁻⁶ C, and a separation of 0.20 m, so r = 0.10 m from each charge to the midpoint.
• Formula: E = E₁ + E₂, with each term E = kQ/r².
• Calculation: E₁ = E₂ = (8.99 × 10⁹ × 1.0 × 10⁻⁶) / (0.10)² = 8.99 × 10⁵ N/C. Both fields point toward the negative charge, so the magnitudes add.
• Answer: E = 1.8 × 10⁶ N/C, directed toward the negative charge.

How Does the Electric Field Compare With Electric Force and Electric Potential?

Electric field is a vector measured in N/C, electric force is a vector measured in N, and electric potential is a scalar measured in V. Table 3 compares 4 electric quantities for a point charge Q and a test charge q at distance r.

Table 3: Electric quantities for a point charge Q at distance r

The quantities connect through F = qE, U = qV, and E = −dV/dr.

How Does the Electric Field Formula Compare With the Gravitational Field Formula?

The electric field E = kQ/r² and the gravitational field g = GM/r² both follow an inverse-square law. The 3 main differences are:

• Charges have 2 signs, which produce attraction or repulsion. Masses have 1 sign, which produces attraction only.
• The Coulomb constant is k = 8.99 × 10⁹ N·m²/C². The gravitational constant is G = 6.674 × 10⁻¹¹ N·m²/kg² (CODATA).
• The electric force between 2 protons exceeds the gravitational force between them by a factor near 1.2 × 10³⁶.

What Are the Applications of the Electric Field Formula?

The electric field formula applies to capacitors, cathode ray tubes, particle accelerators, electrostatic precipitators, and inkjet printers. Parallel-plate capacitors store energy in the uniform field E = V/d. Electrostatic precipitators remove particles from exhaust gas with strong fields.

Research and reference data document 3 applications of electric field calculations:

• Millikan oil drop experiment: Robert Millikan balanced charged oil drops between plates, where qE = mg, and measured the elementary charge between 1909 and 1913. The current defined value is 1.602176634 × 10⁻¹⁹ C.
• Thomson cathode ray experiment: J. J. Thomson deflected cathode rays with an electric field in 1897 and determined the charge-to-mass ratio of the electron.
• Dielectric breakdown of air: Physics references, including OpenStax University Physics Volume 2, list the dielectric strength of dry air near 3 × 10⁶ V/m, which is the field value where sparks form.

What Mistakes Occur Most Often With the Electric Field Formula?

5 mistakes occur most often with the electric field formula:

1. Using centimeters or microcoulombs without conversion to meters and coulombs.
2. Ignoring direction when adding fields from several charges.
3. Applying E = kQ/r² to extended objects at short distances.
4. Confusing the source charge Q with the test charge q.
5. Mixing electric field E with electric potential V.

How Can an Electric Field Be Calculated With a Calculator or Spreadsheet?

Enter 8.99E9 × Q ÷ r² into a scientific calculator, or use =8.99E9*Q/r^2 in a spreadsheet. The formula =8.99E9*2E-6/0.5^2 returns 71920 N/C for Q = 2 μC and r = 0.5 m. For full precision, replace 8.99E9 with 1/(4*PI()*8.8541878188E-12).

What Do People Also Ask About the Electric Field Formula?

What Is the Formula for Electric Field Strength?

Electric field strength equals force divided by charge, E = F/q.

What Is the Electric Field Formula for a Point Charge?

The point-charge formula is E = kQ/r², where k = 8.99 × 10⁹ N·m²/C².

Is the Electric Field a Vector or a Scalar?

The electric field is a vector quantity with magnitude and direction.

What Is the Unit of Electric Field?

The unit of electric field is N/C, which equals V/m.

Does the Electric Field Depend on the Test Charge?

The electric field does not depend on the test charge, because the source charge and the distance determine it.

What Is the Electric Field Inside a Conductor?

The electric field inside a conductor is zero in electrostatic equilibrium.

What Is the Electric Field at the Center of a Charged Ring?

The field at the center of a uniformly charged ring is zero, because opposite sides cancel.

How Does Distance Change the Electric Field of a Point Charge?

Electric field strength decreases with the square of the distance, so doubling r reduces E to one fourth.

What Is the Difference Between Electric Field and Electric Potential?

Electric field is a vector measured in N/C, and electric potential is a scalar measured in V.

What Is the Value of ε₀?

The vacuum permittivity ε₀ is 8.8541878188 × 10⁻¹² F/m, according to NIST CODATA 2022.

What Are the Key Takeaways About the Electric Field Formula?

The electric field formula is E = F/q, and the point-charge form is E = kQ/r². The 6 key takeaways are:

• Use E = F/q to define electric field strength in N/C.
• Apply E = kQ/r² to point charges and spheres.
• Choose E = V/d for uniform fields between parallel plates.
• Add fields from several charges as vectors.
• Convert all quantities to SI units before substituting.
• Determine the direction from the sign of the source charge.

Maurice Cotterell

MEd., BSc. QTS/OCT – Physics Teacher at Davenant Foundation School
The University of the West Indies

My passion as a teacher and leader in education is to contribute to students’ enhanced learning by creation of a positive teaching/learning environment, where they are motivated to develop strong critical thinking, analytical and problem-solving skills.
As acting Vice-Principal and head of Physics, I have amassed five years of experience in leading meetings, implementing strategies to maintain discipline and school ethos, liaise with stakeholders in planning for future development of the institution; successfully negotiated the implementation of various projects for student development as well as supervising effective execution of such. For the past 19 years, I have incorporated discovery learning within the classroom through practical activities, problem solving and cooperative learning and have realized that students retention in Physics and Mathematics have increased significantly.

I am always cognizant that students have different learning styles, and as such teaching strategies during planning and delivery must take this crucial factor into consideration and be tailored to meet such differences.