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18.1 Electric fields and field lines

An electric field is a region in which a charge experiences a force; field strength E is force per unit positive charge.

Use a small positive test charge to define direction, then calculate force on any charge with its sign included.

Between parallel charged plates, a positive charge feels force in the field direction and a negative charge opposite it.

The field exists without placing a test charge, and field direction is not always the same as force on a negative charge.

Electric field strength satisfies E=F/q, so the force on charge q is F=qE; direction reverses for negative q.

Use E as a vector and q with sign. In a uniform field, force and acceleration are constant if q and mass are constant.

A −2 μC charge in a 3.0×10⁴ N C⁻¹ field experiences 0.060 N opposite the field direction.

E is not force itself and does not depend on the test charge used to measure it.

Electric field lines point in the direction of force on a positive test charge; closer lines indicate stronger field qualitatively.

Lines start on positive charge and end on negative charge or infinity, do not cross, and are perpendicular to conductor surfaces in electrostatic equilibrium.

Parallel equally spaced lines represent a uniform field between large oppositely charged plates.

A negative charge moves opposite to the arrow direction, and field-line density is a model rather than an exact numerical scale.

18.2 Uniform electric fields

In a uniform field, field strength magnitude is E=∆V/∆d, where ∆V is potential difference across perpendicular separation ∆d.

Use metres and volts, and remember field direction points from higher potential toward lower potential for a positive test charge.

A 600 V difference across 0.020 m gives E=3.0×10⁴ N C⁻¹.

The relation is for a uniform field; in a point-charge field strength changes with distance.

$F=qE, acceleration magnitude a=|q|E/m$

Charge	Force and acceleration direction
q>0	along the electric field
q<0	opposite the electric field

While the particle is in a uniform field, force and acceleration are constant. The velocity component parallel to the force changes uniformly; a perpendicular velocity component remains constant if other forces are negligible.

Region	Motion for entry perpendicular to E
between the plates	parabolic curved path toward the force direction
beyond the plates	no electric force, so a straight line tangent to the path at exit

At the same E, acceleration depends on |q|/m. A helium nucleus has twice a proton's charge but about four times its mass, so its electric acceleration and perpendicular velocity gain are half as large for the same transit time.

Do not continue curving the trajectory after the particle leaves the field. Also compare charge-to-mass ratio, not charge alone; a negative particle curves opposite the field arrows.

18.3 Electric force between point charges

For a spherical conductor in electrostatic equilibrium, excess charge resides on the surface and the external field acts as if total charge Q were at the centre.

Use centre distance r and remember the field inside the conductor is zero in electrostatic equilibrium.

A charged metal sphere produces an external field decreasing as 1/r², even though the charge is spread over its surface.

The point-charge equivalence does not describe the field inside the conductor or an irregular charged object without symmetry.

Coulomb's law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of their separation.

$force magnitude F=|Q1Q2|/(4πε0r²), where ε0=8.85×10⁻¹² F m⁻¹$

Charges	Direction of the two equal and opposite forces
same sign	repel along the joining line
opposite signs	attract along the joining line

For +3.0 nC and -5.0 nC separated by 0.040 m, F=(3.0×10⁻⁹)(5.0×10⁻⁹)/[4π(8.85×10⁻¹²)(0.040)²]=8.43×10⁻⁵ N. The force is attractive.

$Q doubled → F doubled; r doubled → F divided by 4$

Use centre-to-centre separation in metres and square it. Force magnitude is positive; charge signs choose attraction or repulsion. With several charges, calculate each force vector and add vectors.

18.4 Electric field of a point charge

$field magnitude E=|Q|/(4πε0r²), ε0=8.85×10⁻¹² F m⁻¹$

Source charge Q	Field direction at the point
positive	radially away from Q
negative	radially toward Q

The field belongs to the source charge and is independent of any test charge. For a charged spherical conductor and an external point, use the sphere's total charge at its centre and measure r from the centre.

For Q=83 pC and r=0.620 m, E=(83×10⁻¹²)/[4π(8.85×10⁻¹²)(0.620)²]=1.94 N C⁻¹, directed outward because Q is positive.

$For two source contributions, E1/E2=(|Q1|/|Q2|)(r2²/r1²); the common factor 1/(4πε0) cancels.$

Superposition step	Action
1	calculate each field magnitude at the point
2	draw each direction away from + or toward -
3	add vectors; add collinear same directions and subtract opposite directions

Do not insert a test charge into the field formula—that belongs in F=qE. Field strength falls as 1/r², and multiple fields must be added as vectors, not automatically as magnitudes.

18.5 Electric potential

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.

$Along coordinate x: E_x=-dV/dx or locally E_x≈-ΔV/Δx$

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.

$V=Q/(4πε0r), with V=0 at infinity$

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.

$V_total=Σ Q_i/(4πε0r_i): calculate each signed contribution at the same point, then add algebraically.$

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.

$For source Q, V=Q/(4πε0r). Placing charge q there gives E_P=qV=Qq/(4πε0r), with E_P=0 at infinite separation.$

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.

$For a charge q moving between points: ΔE_P=qΔV=q(V_f-V_i). If only electric forces act, ΔE_K=-ΔE_P.$

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.

18. Electric fields

Exam analysis

Most tested topics

18.1 Electric fields and field lines

An electric field is a region where a charge experiences electric force

The electric force on charge q in field E is F=qE

Electric field lines show force direction and relative strength

18.2 Uniform electric fields

A uniform electric field has strength E=∆V/∆d between equipotential planes

A uniform electric field gives a charged particle constant acceleration

18.3 Electric force between point charges

Outside a charged spherical conductor, its field is equivalent to a point charge at the centre

Use Coulomb's inverse-square law for two point charges

18.4 Electric field of a point charge

Calculate and combine inverse-square electric fields from point charges

18.5 Electric potential

Electric potential is work done per unit positive charge at a point

Electric field is the negative local gradient of electric potential

Point-charge potentials retain charge sign and add as scalars

Electric potential gives the signed potential energy of two charges