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Electric Force Calculator

Coulomb's law, electric field, potential and potential energy with step-by-step solutions
Find the electric force between +3.0 microC and -5.0 microC separated by 0.20 m
Find the electric field and potential 5.0 cm from a 2.0 nC point charge
Find the potential energy of +4.0 microC and +6.0 microC 0.30 m apart
Two equal charges repel with 0.50 N at 0.10 m. What is each charge?

Coulomb's Law

Two point charges attract or repel along the line joining them with a force of magnitude

F=k∣q1q2∣r2,k=14πε0=8.988×109 N\cdotpm2/C2F = k\frac{|q_1 q_2|}{r^2}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 8.988 \times 10^{9}\ \text{N·m}^2\text{/C}^2

  • FF — force, newtons (N)
  • q1,q2q_1, q_2 — charges, coulombs (C); note 1 μC = 10−610^{-6} C and 1 nC = 10−910^{-9} C
  • rr — separation between the charges, metres (m)
  • ε0=8.854×10−12\varepsilon_0 = 8.854 \times 10^{-12} C²/(N·m²), the permittivity of free space

The sign convention that actually works: take the magnitude from the absolute values, then decide the direction physically — like charges repel, unlike charges attract. Charge is quantised in units of e=1.602×10−19e = 1.602 \times 10^{-19} C, so a net charge is always an integer multiple of ee.

The assumption people forget: this is an inverse-square law for point charges or uniform spheres in vacuum, and it gives the force from one other charge. With three or more charges, add the individual forces as vectors.

Field, Potential and Potential Energy

The same 1/r21/r^2 structure reappears in every electrostatic quantity, so it is worth keeping the four straight:

E=Fq0=k∣Q∣r2,V=kQr,U=kq1q2rE = \frac{F}{q_0} = k\frac{|Q|}{r^2}, \qquad V = k\frac{Q}{r}, \qquad U = k\frac{q_1 q_2}{r}

  • EE — electric field, N/C (equivalently V/m); a vector
  • VV — electric potential, volts (V = J/C); a scalar, and it keeps the sign of QQ
  • UU — potential energy of the pair, joules (J)

Notice the pattern: force and field go as 1/r21/r^2, energy and potential go as 1/r1/r. The links between them are F=q0EF = q_0 E and U=q0VU = q_0 V, and the work done moving a charge through a potential difference is

W=q ΔVW = q\,\Delta V

in joules.

When it applies: the kQ/rk Q/r forms assume the zero of potential is at infinity, which is the standard choice for isolated point charges but not for parallel plates or circuits.

Common Mistakes to Avoid

  • Leaving charges in microcoulombs — convert to coulombs first. Forgetting the 10−610^{-6} on both charges is a factor of 101210^{12} error.
  • Using centimetres for rr — the separation must be in metres before it is squared.
  • Squaring rr in the potential — VV and UU go as 1/r1/r, only FF and EE go as 1/r21/r^2.
  • Adding fields as scalars — EE is a vector. Two fields of 100 N/C at right angles give 141 N/C, not 200 N/C.
  • Dropping the sign in UU — an attracting pair has negative potential energy, and that minus sign is what makes the pair bound.
  • Assuming zero field means zero potential — midway between two equal positive charges the field cancels but the potential is at a maximum.

Examples

Step 1: Convert: q1=3.0×10−6 Cq_1 = 3.0 \times 10^{-6}\ \text{C}, q2=5.0×10−6 Cq_2 = 5.0 \times 10^{-6}\ \text{C} (magnitude)
Step 2: F=k∣q1q2∣/r2F = k|q_1 q_2|/r^2 with k=8.988×109 N\cdotpm2/C2k = 8.988 \times 10^{9}\ \text{N·m}^2\text{/C}^2
Step 3: Numerator: (8.988×109)(3.0×10−6)(5.0×10−6)=0.1348 N\cdotpm2(8.988 \times 10^{9})(3.0 \times 10^{-6})(5.0 \times 10^{-6}) = 0.1348\ \text{N·m}^2
Step 4: Denominator: r2=(0.20 m)2=0.040 m2r^2 = (0.20\ \text{m})^2 = 0.040\ \text{m}^2
Step 5: F=(0.1348 N\cdotpm2)÷(0.040 m2)=3.37 NF = (0.1348\ \text{N·m}^2) \div (0.040\ \text{m}^2) = 3.37\ \text{N}
Answer: F≈3.4F \approx 3.4 N, attractive (the charges have opposite signs)

Step 1: Convert: Q=2.0×10−9 CQ = 2.0 \times 10^{-9}\ \text{C}, r=0.050 mr = 0.050\ \text{m}
Step 2: kQ=(8.988×109 N\cdotpm2/C2)(2.0×10−9 C)=17.98 N\cdotpm2/CkQ = (8.988 \times 10^{9}\ \text{N·m}^2\text{/C}^2)(2.0 \times 10^{-9}\ \text{C}) = 17.98\ \text{N·m}^2\text{/C}
Step 3: E=kQ/r2=(17.98 N\cdotpm2/C)÷(0.0025 m2)=7.19×103 N/CE = kQ/r^2 = (17.98\ \text{N·m}^2\text{/C}) \div (0.0025\ \text{m}^2) = 7.19 \times 10^{3}\ \text{N/C}
Step 4: V=kQ/r=(17.98 N\cdotpm2/C)÷(0.050 m)=360 VV = kQ/r = (17.98\ \text{N·m}^2\text{/C}) \div (0.050\ \text{m}) = 360\ \text{V}
Answer: E≈7.2×103E \approx 7.2 \times 10^{3} N/C pointing away from the charge; V≈360V \approx 360 V

Step 1: U=kq1q2/rU = kq_1q_2/r, keeping both signs (both positive here)
Step 2: q1q2=(4.0×10−6 C)(6.0×10−6 C)=2.4×10−11 C2q_1 q_2 = (4.0 \times 10^{-6}\ \text{C})(6.0 \times 10^{-6}\ \text{C}) = 2.4 \times 10^{-11}\ \text{C}^2
Step 3: kq1q2=(8.988×109)(2.4×10−11)=0.2157 N\cdotpm2k q_1 q_2 = (8.988 \times 10^{9})(2.4 \times 10^{-11}) = 0.2157\ \text{N·m}^2
Step 4: U=(0.2157 N\cdotpm2)÷(0.30 m)=0.719 JU = (0.2157\ \text{N·m}^2) \div (0.30\ \text{m}) = 0.719\ \text{J}
Answer: U≈+0.72U \approx +0.72 J — positive, so work had to be done to bring them together

Frequently Asked Questions

Coulomb's law, F = k|q₁q₂|/r², with k = 8.988 × 10⁹ N·m²/C², charges in coulombs and separation in metres. The result is a force in newtons, repulsive for like charges and attractive for unlike ones.

V = kQ/r, in volts, with the zero of potential taken at infinity. Unlike the field, potential is a scalar and keeps the sign of Q, so several charges combine by simple addition rather than by vector addition.

The field E = kQ/r² is a property of the source charge alone, measured in N/C, and exists whether or not anything is there to feel it. The force on a test charge placed in that field is F = q₀E, in newtons, so it depends on both charges.

Because k is 8.99 × 10⁹ while G is 6.67 × 10⁻¹¹. Two protons repel electrically about 10³⁶ times more strongly than they attract gravitationally. Gravity only wins on astronomical scales because bulk matter is electrically neutral.

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