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Electrostatics questions are short, which makes them easy to rush. The traps are almost always the same: adding the wrong kind of quantity, losing a sign, or assuming zero field means zero potential. Here are four mistakes, each with numbers you can check. We use k ≈ 9 × 10⁹ N m²/C² (OpenStax gives 8.99 × 10⁹).
Mistake 1: Adding fields and potentials the same way
OpenStax University Physics states the rule directly: for potential, "you add the individual voltages as numbers", while for field "you must add the individual fields as vectors, taking magnitude and direction into account" (OpenStax, calculations of electric potential). The same text notes that V is closely associated with energy, a scalar, whereas E is closely associated with force, a vector. NCERT's Class 12 chapter on charges and fields describes the field of several charges as a vector sum by the superposition principle.
Worked example A. Two +2 μC charges sit at x = −0.1 m and x = +0.1 m. At the midpoint:
Each field has magnitude kq/r² = (9 × 10⁹)(2 × 10⁻⁶)/(0.1)² = 1.8 × 10⁶ N/C, pointing in opposite directions. Net E = 0.
Each potential is kq/r = (9 × 10⁹)(2 × 10⁻⁶)/0.1 = 1.8 × 10⁵ V. Net V = 3.6 × 10⁵ V.
Worked example B. Make the right-hand charge −2 μC. Now the potentials cancel, so V = 0, but both fields point towards the negative charge, so E = 3.6 × 10⁶ N/C in the +x direction.
The lesson: E = 0 does not mean V = 0, and V = 0 does not mean E = 0.
Mistake 2: Dropping the sign of the charge
For potential and potential energy, the sign of the charge goes straight into the formula. For fields, it is usually safer to compute magnitudes and take directions from a diagram. Mixing the two methods in one line is where signs disappear.
NCERT's chapter on electrostatic potential gives the potential energy of two charges as U = q₁q₂/(4πε₀r₁₂), which carries the signs of both charges.
Worked example. Charges of +3 μC and −3 μC are 0.3 m apart. U = (9 × 10⁹)(3 × 10⁻⁶)(−3 × 10⁻⁶)/0.3 = −0.27 J. The negative sign is the physics: you would have to do 0.27 J of work to pull them apart to infinity. Writing +0.27 J reverses the meaning.
Mistake 3: Assuming the potential is zero inside a conductor
NCERT lists two separate facts: "Inside a conductor, electrostatic field is zero" and "Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface" (NCERT Class 12, Chapter 2). Constant is not the same as zero.
Worked example. A conducting sphere of radius 0.1 m carries 1 nC.
At the surface, V = kQ/R = (9 × 10⁹)(10⁻⁹)/0.1 = 90 V.
Everywhere inside, V = 90 V too, and E = 0.
Just outside, E = kQ/R² = 900 N/C.
A graph of V against distance from the centre is flat at 90 V up to the surface, then falls off as 1/r. A graph of E is zero inside, jumps at the surface, then falls as 1/r².
Mistake 4: Getting 'downhill' wrong for negative charges
NCERT states that the "Electric field is in the direction in which the potential decreases" (NCERT Class 12, Chapter 2). OpenStax defines potential as potential energy per unit charge, V = U/q (OpenStax, electric potential), so the change in potential energy is ΔU = qΔV.
Worked example. An electron moves from a point at 0 V to a point at +100 V. ΔU = (−1.6 × 10⁻¹⁹)(100) = −1.6 × 10⁻¹⁷ J. Its potential energy falls, so this is the direction it moves on its own, gaining 100 eV of kinetic energy. Positive charges move towards lower potential; negative charges move towards higher potential.
Our guide to reading electric and magnetic field lines covers the common misreadings of field lines and equipotentials, and the Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
A quick checklist for any electrostatics answer
Is this quantity a scalar (V, U) or a vector (E, F)? Add accordingly.
Did the sign of every charge go into V and U?
Did every field get a direction from the diagram?
Inside a conductor: E = 0, V = surface value.
Does the sign of ΔU match which way the charge would move on its own?
If you keep losing marks to the same step, compare your working line by line with the Fermi Problem Library, which has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. Many of these errors are also vector slips; see vectors in JEE physics: common errors.
