Physics · Electric Charges And Fields · NEET
No. This is the key idea of superposition. The force on q1 due to q2 is exactly the Coulomb force k·q1·q2/r12^2, whether or not q3, q4, ... are present. Nearby charges do NOT weaken or strengthen a pair force. They only add their OWN separate forces on q1. NCERT states this clearly: the force on q1 due to q2 is unaffected by the presence of other charges.
Because force has direction. F1 from a charge on the left and F2 from a charge above point in different directions. Adding numbers only works if the forces are along the same line. For charges at angles you must break each force into x and y components, add components, then combine: F_net = sqrt(Fx^2 + Fy^2). For NEET, always draw the arrows first.
Not exactly, and NEET can test this. Vector addition is only the second half. Superposition says TWO things: (1) each pair force is unaffected by other charges, and (2) there are no extra three-body or four-body forces that appear only when many charges are present. Once you accept (1) and (2), you then use ordinary vector addition to get the total.
Find the force from each other charge separately using F = k·q·q'/r^2, mark its direction (attraction points toward the other charge, repulsion points away). Then resolve into components and add. Symmetry is your shortcut: equal charges placed symmetrically often give forces that cancel, leaving a simple net direction. At the centre of a square with 4 equal like charges, the net force is zero by symmetry.
If the two outer charges are equal and like-signed, they push (or pull) the middle charge with equal and opposite forces, so the net force is zero. But note: this is an unstable equilibrium along the line and the field there can still be non-zero for a test charge off-axis. Superposition gives zero net force here purely because the two equal forces cancel.
Two point charges A and B, having charges +Q and -Q respectively, are placed 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 are separated by a distance, each carrying charge q, with force of repulsion F between them. A third identical uncharged sphere is touched to A first, then to B, and finally removed. The new force of repulsion between A and B is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It states that the total electric force on a charge is the vector sum of the individual Coulomb forces due to each of the other charges, and each pair force is independent of the other charges present.
F_net = F1 + F2 + F3 + ... (added as vectors), where each Fi = k·q1·qi/ri^2 points along the line joining the pair, with k = 1/(4·pi·epsilon0) = 9 x 10^9 N·m^2/C^2.
Yes. The net electric field at a point is the vector sum of the fields due to each charge: E_net = E1 + E2 + ... The same principle applies to force, field, and potential.
When the vector sum of all individual Coulomb forces cancels. This often happens due to symmetry, for example at the centre of a square with four equal like charges, or midway between two equal like charges.
The principle itself holds in any medium. In a uniform medium you replace k by k/K (K = dielectric constant), but each pair force is still computed independently and then added as vectors.
