Electric Field and Potential
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Direct answer
The electric field is force per unit positive test charge, E = F/q; the potential is work per unit charge in bringing a test charge from infinity, V = k q/r for a point charge, with k = 1/(4 pi epsilon_0) = 9 × 10^9 in SI units. They are linked by E = −dV/dr: the field points down the steepest fall of potential. Fields add as vectors, potentials as scalars — a distinction JEE exploits constantly.
What you must remember
- Point charge: E = k q/r^2 outward from positive charges; V = k q/r, negative for negative charges; pair energy U = k q1 q2/r.
- Uniform field between parallel plates: E = V/d; work in moving charge q is W = q (V_final − V_initial), independent of path.
- Dipole (p = charge × separation): axial field 2 k p/r^3, equatorial k p/r^3; potential k p/r^2 on the axis, zero on the equatorial plane.
- Dipole in a uniform field: torque = p E sin(theta), energy U = −p E cos(theta); stable at theta = 0, unstable at 180 degrees; net force zero.
- Conductors: zero field inside and inside cavities (shielding), constant potential throughout, charge on the outer surface.
- Charged spherical shell: E = 0 and V = k q/R (constant) everywhere inside; outside both behave as for a point charge at the centre. Inside a uniformly charged non-conducting sphere, E grows linearly with r.
- Zero field does not imply zero potential, nor the reverse: between equal like charges E vanishes at the midpoint while V does not; on the dipole's equatorial plane V is zero while E is not.
Common confusion
The persistent error is sign bookkeeping — fields point inward for negative charges, their potentials are negative, and the dipole energy −p E cos(theta) carries its own sign; magnitudes come out right while answers stay wrong. The second classic slip is summing fields and potentials the same way: potential adds algebraically, field only after resolving into components. Remember too that the zero of potential is a reference choice; only differences (and the field, their gradient) are physical.
Exam-focused takeaway
JEE Main tests field and potential of simple charge arrays, a charge released in a uniform field, work done in moves, and dipole torque as single-correct and numerical-value questions. JEE Advanced prefers equilibrium of an inserted charge, field as the slope of a given V(x) graph, angular dipole problems, Gauss-law crossover for spheres and shells, and the potential energy of three- or four-charge systems. Decide per quantity whether you are adding vectors or scalars, and fix signs at the start.
Frequently asked questions
Can E be zero where V is not, and the reverse?
Yes — at the midpoint between equal like charges, E = 0 while V is finite; on a dipole's equatorial plane, V = 0 while E is finite. Field measures how fast potential changes, not its value.
How are E and V related?
E = −dV/dr: the field component along any direction is minus the potential's rate of change in that direction.
Why is the field zero inside a conductor?
Free electrons rearrange until their own field cancels the applied one inside — the basis of electrostatic shielding.
Why is the dipole's aligned state its stable equilibrium?
U = −p E cos(theta) is minimum at theta = 0; the torque p E sin(theta) restores any small rotation, whereas 180 degrees is unstable.
What are E and V inside a charged shell?
E = 0 everywhere inside, while V stays constant at k q/R — constant, not zero.