Physics · Electric Charges And Fields · NEET
Step 1: Write the formula F = k q1 q2 / r squared. Step 2: Convert every charge to coulombs (1 microC = 10^-6 C, 1 nC = 10^-9 C) and the distance to metres (1 cm = 0.01 m). Step 3: Put k = 9 x 10^9. Step 4: Multiply the two charge values and k on top. Step 5: Square the distance and divide. Step 6: Write the answer in newtons. Use only magnitudes to get the size of the force; decide attraction or repulsion from the signs separately (like charges repel, unlike attract).
They are the same number. k = 1/(4 pi epsilon0) = 9 x 10^9 N m^2/C^2 (more precisely 8.99 x 10^9). In NEET numericals always use 9 x 10^9 unless the question gives you epsilon0 = 8.85 x 10^-12 and asks you to derive k. Pick whichever form matches the data given so you do not do extra arithmetic.
Yes. The r in the denominator is squared, so F = k q1 q2 / r^2, not / r. This is the most common silent mistake. If r = 2 cm = 0.02 m, the denominator is (0.02)^2 = 4 x 10^-4, not 0.02. Because of the square, doubling the distance makes the force one-fourth, and halving it makes the force four times larger.
It is almost always a unit slip. microC is 10^-6, nanoC is 10^-9, and cm is 10^-2 m. Two charges in microC give 10^-6 x 10^-6 = 10^-12 on top; a distance in cm gives 10^-2 squared = 10^-4 on the bottom. Convert first, keep the powers of ten in a separate line, and combine them at the end so nothing is lost.
Coulomb's law only gives the force between one pair. For many charges, calculate the force from each other charge on your chosen charge separately, then add them as vectors (superposition). If the forces are along the same line, add or subtract sizes; if they are at an angle, use components. This concept is covered in Superposition Principle.
Two point charges A and B, having charges +Q and -Q respectively, are placed at a certain distance apart and the force acting between them is F. If 25% of the charge of A is transferred to B, then the force between the charges becomes
Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B, and finally removed from both. The new force of repulsion between spheres A and B is best given as:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F = k q1 q2 / r^2, where k = 9 x 10^9 N m^2/C^2, q1 and q2 are the charge magnitudes in coulombs, and r is the separation in metres. The force acts along the line joining the two charges.
Charges in coulombs (C), distance in metres (m), and force comes out in newtons (N). Convert microC (10^-6), nanoC (10^-9), and cm (10^-2 m) before substituting.
Because force depends on 1/r^2, doubling the distance makes the force one-fourth of its original value. Halving the distance makes the force four times larger.
Force is directly proportional to each charge, so doubling one charge doubles the force. Doubling both charges makes the force four times larger.
No. Use only magnitudes to find the size of the force. The signs tell you the direction only: like charges (same sign) repel, unlike charges attract.
