9478 / 2027
Electric Fields overview

Lesson 3 of 7

Field from a potential gradient

  • I can recall and apply electric field strength as the negative local potential gradient.Syllabus

    Syllabus 9478, 14(f). Recall and apply electric field strength as the negative local potential gradient.

Make a guess. You are not marked.

Along a line away from an isolated positive point charge, how does the electric potential V vary with distance r?

  1. It falls from large positive values towards zero, steeply close to the charge.
  2. It falls in a straight line and reaches zero at a fixed distance from the charge.
  3. It rises towards zero from large negative values close to the charge.
Show the answer

It falls from large positive values towards zero, steeply close to the charge.

For a point charge V = Q / (4 pi epsilon0 r). It is positive for a positive charge, falls as 1/r and approaches zero at large distance, so the curve is steepest near the charge.

Electric field follows the local rate of decrease of potential. A potential graph's height, a chord gradient and a tangent gradient describe different quantities.

Use the negative local gradient

Along a chosen coordinate x, the field component is Ex = -dV/dx. For a point source with outward radial coordinate r:

Er = -dV/dr

A small radial movement dr has field work qErdr = -q dV, giving this relation. The field points towards decreasing potential. A negative charge's force points oppositely to the field.

The units agree because 1 V/m = 1 J/(C m) = 1 N/C. Field is a potential gradient, not the potential itself.

For the fixed +4.00 nC source with K = 9.00 × 109 N m2/C2, V(r) = 36/r when r is in metres and V in volts. A negative source of the same magnitude gives -36/r. Both tend to zero at indefinitely large r.

Calculated positive-source curve values, not tangent-guide values
r / mV / V
0.150240
0.200180
0.300120
0.40090
0.60060

Use the slope at the point, not the potential height

These smooth model curves use K|Q| = 36 V m and positive source distance r. The zero reference is infinitely large separation; none of the displayed finite distances is infinity.

The source sign changes the potential curve

Positive and negative point-source potentials approach zero from opposite sidesRadius runs from 0.150 to 0.600 metres on the horizontal coordinate. Potential is in volts, with its zero line as the arrowed horizontal axis. The solid blue positive-source curve is V equals 36 divided by r and decreases from positive 240 to positive sixty volts. The dashed brown negative-source curve is its reflection across zero and increases from negative 240 to negative sixty volts. Both are analytically sampled reciprocal curves with no finite-radius zero. Positive and negative source signs are named by the visible line-style key. The plotted radial domain starts at 0.150 metres, not at the source.0.1500.3000.4500.600-240-1200+120+240V / Vr / m

Solid: Q = +4.00 nC, V = +36/r. Dashed: Q = -4.00 nC, V = -36/r. The vertical label V / V means potential divided by the unit volt.

At r = 0.300 m, take the negative tangent gradient

The positive-source tangent has gradient negative four hundred volts per metreThe blue curve is V equals 36 divided by r between 0.150 and 0.450 metres. Filled circles mark actual curve values at 0.200 metres and 180 volts, 0.300 metres and 120 volts, and 0.400 metres and ninety volts. The brown dashed tangent is V equals 240 minus 400 r. Its hollow square guides are 0.200 metres and 160 volts, and 0.400 metres and eighty volts; these are not curve measurements. The tangent contacts the curve at the filled 0.300-metre point. A right-angle slope guide has positive radial interval 0.200 metres and negative voltage change eighty volts. Thus the local gradient is negative four hundred volts per metre and the outward-positive field component is positive four hundred newtons per coulomb. The chord through the outer two filled curve points would give negative 450 volts per metre and is not the local tangent.0.2000.3000.400080120160200240Δr = +0.200 mΔV-80 VV / Vr / m

Filled circles are curve points; hollow squares are tangent guides. The local slope is -80/0.200 = -400 V/m. Since Er = -dV/dr, the field is 400 N/C outward. A chord through the actual 0.200 and 0.400 m readings has a different slope.

The overview compares opposite source signs. The close view shows the tangent at r = 0.300 m. Hollow guide marks lie on the tangent only; filled marks represent actual curve values.

Worked local field

Read the tangent rather than a chord

The tangent at (0.300 m, 120 V) passes through the line-only guide points (0.200 m, 160 V) and (0.400 m, 80 V).

Tangent gradient = (80 - 160)/(0.400 - 0.200)
= -400 V/m
Er = -(-400) = +400 N/C

The field is outward. Using the actual curve readings at 0.200 and 0.400 m instead gives (90 - 180)/0.200 = -450 V/m. That is a finite-interval chord gradient, not this point's tangent gradient.

Check your understanding

Outward radial distance is positive. The tangent to V(r) at r = 0.300 m passes through the guide points (0.200 m, 160 V) and (0.400 m, 80 V). What is the radial electric field there?

  1. +400 N/C, outward.
  2. -400 N/C, inward.
  3. +120 N/C, because the potential at the contact point is 120 V.
Show the answer

+400 N/C, outward.

The tangent gradient is (80 - 160)/(0.400 - 0.200) = -400 V/m. E_r = -dV/dr therefore gives +400 V/m = +400 N/C, outward.

Equipotentials and zero points

Potential is constant along an equipotential. The electrostatic field is perpendicular to it; moving along it gives ΔV = 0 and therefore zero field work on a charge. A point with zero field does not automatically have zero potential.

Consider fixed sources at x = -0.300 and +0.300 m, with the common zero-potential reference at infinity. At their midpoint, add field vectors but add signed scalar potentials.

Add fields as vectors and potentials as signed scalars

In both source arrangements, P is 0.300 m from each source. The arrows below each arrangement are the component fields evaluated at P, translated for comparison; they are not paths through space. Their common length scale is 85 drawing units per 400 N/C.

Equal positive sources: field cancels

Equal positive sources: field cancels at the midpointThe left source at negative 0.300 metres is positive four nanocoulombs. The right source at positive 0.300 metres is positive four nanocoulombs. P is the centre point at zero. Each individual field at P has magnitude four hundred newtons per coulomb. The left source field points right. The right positive source field points left. The component vectors cancel and no nonzero resultant arrow is drawn. The two positive 120-volt potentials add to positive 240 volts. Labels name the left and right source contributions. Vector arrows are separate comparison rows, not additional source locations. Zero field and zero potential are different conditions.+++4.00 nC+4.00 nCP0.300 m0.300 mFrom left source400 N/CFrom right source400 N/CResultant E = 0V(P) = +120 + 120= +240 V

Opposite sources: potential cancels

Opposite sources: potential cancels at the midpointThe left source at negative 0.300 metres is positive four nanocoulombs. The right source at positive 0.300 metres is negative four nanocoulombs. P is the centre point at zero. Each individual field at P has magnitude four hundred newtons per coulomb. The left source field points right. The right negative source field also points right. The separate resultant vector points right with twice the component length, representing eight hundred newtons per coulomb. The two signed potentials are positive and negative 120 volts, whose sum is zero. Labels name the left and right source contributions. Vector arrows are separate comparison rows, not additional source locations. Zero field and zero potential are different conditions.+-+4.00 nC-4.00 nCP0.300 m0.300 mFrom left source400 N/CFrom right source400 N/CResultant: 800 N/C rightV(P) = +120 - 120= 0 V
For two equal positive sources the midpoint component fields cancel, so there is no nonzero resultant arrow. For the positive-left, negative-right pair, the component fields reinforce while their potentials cancel.
  • Two +4.00 nC sources: fields are +400 and -400 N/C, so E = 0. Potentials are +120 and +120 V, giving V = +240 V.
  • +4.00 nC on the left, -4.00 nC on the right: both fields point right, giving E = +800 N/C. Potentials +120 and -120 V sum to zero.

These results are consistent with E being a local gradient. A function can have a zero gradient at nonzero height, or pass through zero with a nonzero gradient.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

State how the electric field strength at a point is related to the electric potential there. Two parallel plates 2.0 cm apart have a potential difference of 500 V between them. Calculate the field strength between the plates. [3]

Saved on this browser only. Anyone using this browser can see it.