18.5 Electric potential
- Syllabus
- 9702–2028–2029
- Topic
- 18.5
- Level
- A2
Learning objectives
18.5.1Electric potential at a point as the work done per unit positive charge in• define electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point18.5.2The fact that the electric field at a point is equal to the negative of• recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point18.5.3V = Q / (4πε0 r) for the electric potential in the field due to a point• use V = Q / (4πε0 r) for the electric potential in the field due to a point charge18.5.4The concept of electric potential leads to the electric potential energy of• understand how the concept of electric potential leads to the electric potential energy of two point charges and use EP = Qq / (4πε0 r)
Electric potential is work done per unit positive charge at a point
Electric potential V at a point is work done per unit positive test charge bringing it from infinity, with zero potential at infinity.
Potential is scalar, so contributions add algebraically; electric field is a vector and requires direction.
A positive point charge gives positive potential, while a negative source gives negative potential relative to infinity.
Potential is not force per charge—that is field strength—and equal potential does not mean zero field everywhere.
Electric field is the negative local gradient of electric potential
Alongcoordinatex:Ex=−dV/dxorlocallyEx≈−ΔV/Δx
Electric field points in the direction in which electric potential decreases most rapidly. The minus sign converts the signed potential gradient into field direction.
| V–x graph feature at a point | Electric field |
|---|---|
| straight negative gradient | constant positive E_x |
| straight positive gradient | constant negative E_x |
| horizontal tangent | E_x=0 |
| steeper tangent magnitude | larger |
Between plates, V falls linearly from 600 V to 0 V over +0.020 m. dV/dx=-3.0×10⁴ V m⁻¹, so E_x=+3.0×10⁴ N C⁻¹.
For a curved V–x graph, draw a tangent at the requested point and use its gradient. A secant across a wide interval gives only an average field, not the exact local field.
A high value of V does not imply a large E: field depends on how rapidly V changes with position. Always retain the minus sign when direction is required.
Point-charge potentials retain charge sign and add as scalars
V=Q/(4πε0r),withV=0atinfinity
Potential is scalar and keeps the source charge sign: positive Q gives positive V and negative Q gives negative V. It varies as 1/r, not 1/r².
At r=0.030 m from Q=+2.0 nC, V=(2.0×10⁻⁹)/[4π(8.85×10⁻¹²)(0.030)]=+599 V.
Vtotal=ΣQi/(4πε0ri):calculateeachsignedcontributionatthesamepoint,thenaddalgebraically.
A point 0.020 m from +3.0 nC and 0.060 m from -3.0 nC has V_total=(1/4πε0)[3.0×10⁻⁹/0.020-3.0×10⁻⁹/0.060]=+899 V.
Do not use |Q| or vector directions when adding potential. V can be zero because signed scalar contributions cancel even when the resultant electric field is not zero.
Electric potential gives the signed potential energy of two charges
ForsourceQ,V=Q/(4πε0r).PlacingchargeqtheregivesEP=qV=Qq/(4πε0r),withEP=0atinfiniteseparation.
| Pair | Sign of Qq and E_P | Physical meaning relative to infinity |
|---|---|---|
| like charges | positive | external work is required to bring them closer |
| unlike charges | negative | energy is released as attraction brings them closer |
For a proton and electron separated by 5.3×10⁻¹¹ m, E_P=-(1.60×10⁻¹⁹)²/[4π(8.85×10⁻¹²)(5.3×10⁻¹¹)]=-4.35×10⁻¹⁸ J.
Forachargeqmovingbetweenpoints:ΔEP=qΔV=q(Vf−Vi).Ifonlyelectricforcesact,ΔEK=−ΔEP.
A positive particle moving through a potential drop loses E_P and gains the same kinetic energy. A negative particle reverses the sign relation because q<0.
Keep both charge signs in Qq: potential energy may be negative. Use E_P=qV for energy at one point and ΔE_P=qΔV for a move between two points; these are not interchangeable.