According to Coulomb's law, the electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ is proportional to
A$\frac{q_1 q_2}{r^2}$
B$\frac{q_1 q_2}{r}$
C$\frac{q_1 + q_2}{r^2}$
D$q_1 q_2 \cdot r^2$
Answer & Solution
Correct answer: A. $\frac{q_1 q_2}{r^2}$
1. Coulomb measured the force between two charged spheres with a torsion balance (1785).
2. He found the force grows with the PRODUCT of the two charges: double either charge, the force doubles.
3. He also found the force falls with the SQUARE of the separation: double the distance, force drops to one fourth.
4. Combining both: $F = k\frac{q_1 q_2}{r^2}$. Option B misses the square, option C adds charges instead of multiplying, option D grows with distance.
_Source: syllabus Physics Ch 1 "Electric Charges and Fields", §1.6 Coulomb's Law_
Keep practising Electric Charges and Fields
Browse all Electric Charges and Fields questions for NEET UG, or explore more Physics topics.
Related questions
NEET · Physics · Electric Charges & Fields · Force on Dipole in FieldElectric potential is a:True or false: electrostaticsThe work done in bringing a charge q from infinity to a point where the potential is V isA parallel-plate capacitor in vacuum has plate area A and separation d. Its capacitance isOn an equipotential surface, the work done in moving a test charge between any two points Electric field at any point inside a hollow charged conductor isAn electric dipole of moment p is placed in a uniform field E. The work done to rotate it