Q.A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of $9.6\times10^{-19}$ C falls between the plates. Find

The electric field is the trick that lets us stop talking about charges pulling on charges directly, and instead say that a charge fills the space around it with a condition — a field — that any other charge then simply responds to. Bring a tiny positive test charge q₀ to a point and measure the force F it feels; the field there is

E = F/q₀ — the force per unit positive charge, in N·C⁻¹.

Dividing out q₀ is the whole point: E describes the space, not the visitor. Once you know E at a point, the force on any charge q placed there is just F = qE — along E if q is positive, opposite if negative.

Field of a point charge. Put a charge Q at the origin. The field it makes a distance r away has magnitude E = kQ/r² with k = 9 × 10⁹, pointing radially outward from a positive Q and inward toward a negative one. It is Coulomb's law with one q₀ stripped off — same inverse-square falloff.

Superposition. Fields add as vectors. The net field of many charges is the vector sum E = E₁ + E₂ + …, each term computed as if the others were absent. Resolve into components; never add magnitudes blindly.

Field lines picture this: they start on positive charge, end on negative, never cross (the field has one direction at each point), and crowd together where the field is strong. The tangent gives the field's direction. …