Which expression correctly relates ε0 to Coulomb's constant k?

Which expression correctly relates ε0 to Coulomb's constant k?

• A. ε0 = 4πk
• B. ε0 = k
• C. ε0 = 1/4πk
• D. ε0 = π/k

Answer: (C)

Coulomb’s constant k and the vacuum permittivity ε0 are tied together by how electric fields from a point charge behave in space. In vacuum, Gauss’s law gives the electric field from a point charge as E = q/(4π ε0 r^2). Coulomb’s law expresses the same field as E = k q / r^2. Since these two expressions describe the same physical field, the coefficients must match: k = 1/(4π ε0). Solving for ε0 gives ε0 = 1/(4π k). This is why the relation ε0 = 1/(4π k) is correct. Numerically, ε0 ≈ 8.854×10^-12 F/m and k ≈ 8.987×10^9 N·m^2/C^2, which are consistent with that relationship.

Coulomb’s constant k and the vacuum permittivity ε0 are tied together by how electric fields from a point charge behave in space. In vacuum, Gauss’s law gives the electric field from a point charge as E = q/(4π ε0 r^2). Coulomb’s law expresses the same field as E = k q / r^2. Since these two expressions describe the same physical field, the coefficients must match: k = 1/(4π ε0). Solving for ε0 gives ε0 = 1/(4π k). This is why the relation ε0 = 1/(4π k) is correct. Numerically, ε0 ≈ 8.854×10^-12 F/m and k ≈ 8.987×10^9 N·m^2/C^2, which are consistent with that relationship.