MixedJEE Physics · Original learning card10 original chapter questions

Electric Field of Discrete Charges

For point sources, E(r) = sum_i k q_i (r-r_i)/|r-r_i|^3. The field direction is the force direction on a positive test charge.

Why this shows up in the exam

Field at a square corner · Field at a circle centre · Finding charge ratios from field direction

Learn the idea

Electric field is force per positive test charge and superposes as a vector. A source configuration sets a field at a point even before a test charge is placed there. Symmetry and components usually simplify the vector sum.

🧠 Memory hook: Pretend the test charge is positive, then add field arrows.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • E = F/q₀ — definition in the test-charge limit
  • E = sum_i k q_i R_i/R_i³ — vector field of discrete point charges

How to approach it

  1. 1Draw source-to-field-point vectors
  2. 2Resolve and sum field components
  3. 3Check units N/C and symmetry

Common slip-ups that cost marks

  • •Multiplying by a test charge when only E is asked
  • •Adding scalar field magnitudes
  • •Reversing the field of a negative source

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

Two point charges 1 microC and 2 microC are 1 m apart in vacuum. Take k = 9 x 10^9 SI. Find the force magnitude.

Take a timed JEE Physics sectional mock