Skip to content
Example · Example 2

Q.State Coulomb's law in words and write it as a vector equation for the force F⃗12\vec{F}_{12} exerted by charge q1q_1 on charge q2q_2. Identify the value and SI unit of the Coulomb constant k=1/4πϵ0k = 1/4\pi\epsilon_0.

West Bengal WbchseTextbookSubjectiveImportance★★★★★
5% · 2/42 Questions
✓ Free question

Coulomb's law: the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them, acting along the line joining them.

F⃗21=kq1q2r2r^21,k=14πϵ0\vec{F}_{21} = \frac{kq_1q_2}{r^2}\hat{r}_{21}, \qquad k=\frac{1}{4\pi\epsilon_0}

where r^21\hat{r}_{21} points from q1q_1 to q2q_2. If q1q2>0q_1q_2>0 (same sign) the force is repulsive (along +r^21+\hat{r}_{21}); if q1q2<0q_1q_2<0 (opposite sign) it is attractive (along −r^21-\hat{r}_{21}).

The Coulomb constant k=1/4πϵ0≈9×109 N m2/C2k=1/4\pi\epsilon_0\approx9\times10^9\ \text{N}\,\text{m}^2/\text{C}^2, where ϵ0=8.854×10−12 C2N−1m−2\epsilon_0=8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2} is the permittivity of free space.

✓Final answer

F⃗21=kq1q2r2r^21\vec{F}_{21}=\dfrac{kq_1q_2}{r^2}\hat{r}_{21}; k≈9×109 N m2/C2k\approx9\times10^9\ \text{N}\,\text{m}^2/\text{C}^2.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.