How to Interpret a Graph of Electric Potential

An electric-potential graph normally shows the electric potential $V$ on the vertical axis and position $x$ on the horizontal axis. The most useful fact is that the electric field is related to the slope of this graph:

$$E_x=-\frac{dV}{dx}$$

Thus, the graph does more than show voltage. It tells you the magnitude and direction of the electric field at every position. This interpretation applies to a one-dimensional situation, such as motion along the $x$-axis.

What the slope means

The electric field points in the direction in which electric potential decreases most rapidly. In one dimension:

• If $V$ increases as $x$ increases, then $dV/dx>0$ and $E_x<0$.
• If $V$ decreases as $x$ increases, then $dV/dx<0$ and $E_x>0$.
• If the graph is horizontal, then $dV/dx=0$ and the electric field is zero.
• A steeper graph means a larger electric-field magnitude.

For a graph made of straight-line sections, use the average slope of each section:

$$E_x=-\frac{\Delta V}{\Delta x}$$

The units are equivalent because $1\ \mathrm{V/m}=1\ \mathrm{N/C}$.

The minus sign is essential. Electric field points from higher potential toward lower potential, although a positive charge would naturally accelerate in the direction of the electric field.

Worked example: a straight-line graph

Suppose the potential changes from $20\ \mathrm{V}$ at $x=0$ to $5\ \mathrm{V}$ at $x=3\ \mathrm{m}$. The graph is a straight line, so the field is uniform.

First calculate the slope:

$$\frac{\Delta V}{\Delta x}=\frac{5\ \mathrm{V}-20\ \mathrm{V}}{3\ \mathrm{m}-0\ \mathrm{m}} =-5\ \mathrm{V/m}$$

Then apply the minus sign:

$$E_x=-\frac{\Delta V}{\Delta x}=+5\ \mathrm{V/m}$$

The electric field has magnitude $5\ \mathrm{N/C}$ and points in the positive $x$-direction. The potential decreases as you move toward positive $x$, which is exactly what the positive field direction indicates.

The potential function for this example is

$$V(x)=20\ \mathrm{V}-(5\ \mathrm{V/m})x.$$

Common graph shapes

Horizontal line: zero field

If $V$ is constant over an interval, then

$$\frac{dV}{dx}=0,$$

so the electric field is zero throughout that interval. This occurs inside an ideal conductor in electrostatic equilibrium and can also occur in a field-free region.

A constant potential does not necessarily mean that the potential is zero. A horizontal line at $100\ \mathrm{V}$ still represents zero electric field. Electric potential is defined up to an arbitrary reference value; only potential differences affect the electric field and electrostatic work.

Straight sloping line: uniform field

A straight line has a constant slope. Therefore, it represents a constant electric field. For example,

$$V(x)=V_0-Ex$$

has

$$-\frac{dV}{dx}=E.$$

This is the idealized result between large, parallel charged plates away from their edges.

Curved line: nonuniform field

For a curved potential graph, the field changes with position. You must find the tangent slope at the position of interest:

$$E_x(x)=-\frac{dV}{dx}.$$

A graph can be increasing but becoming less steep. In that case the electric field remains negative, but its magnitude decreases. A graph can also have a local maximum or minimum. At such a smooth turning point, the slope is zero, so the electric field is zero at that point.

Discontinuous potential

In ordinary electrostatics, electric potential is continuous across a finite surface charge, although its derivative—and therefore the normal component of the electric field—can change abruptly. A genuine jump in potential requires an idealized dipole layer or another special model. Therefore, do not automatically interpret every drawn vertical jump as a usual charged surface.

Example: potential from a point charge

For a point charge $Q$, choosing $V=0$ at infinity gives

$$V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}.$$

For a positive charge, $V$ is positive and decreases toward zero as $r$ increases. Its graph is curved rather than linear. Differentiating gives

$$E_r=-\frac{dV}{dr} =\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2},$$

which points radially outward for $Q>0$.

For a negative point charge, the potential is negative. As $r$ increases, the potential rises toward zero, so $dV/dr>0$ and $E_r<0$: the field points inward, toward the charge.

For example, if a positive point charge produces potentials of $90\ \mathrm{V}$ at $r=1\ \mathrm{m}$ and $22.5\ \mathrm{V}$ at $r=2\ \mathrm{m}$, the potential has fallen by a factor of four. This reflects the $1/r$ dependence. The corresponding field falls even faster, as $1/r^2$.

Potential versus distance is not always potential energy

Electric potential $V$ is potential energy per unit charge:

$$V=\frac{U}{q}.$$

A graph of $U$ against position is different from a graph of $V$ against position by a factor of charge. For a particle with charge $q$,

$$U(x)=qV(x),$$

and the force is

$$F_x=-\frac{dU}{dx}=qE_x.$$

A positive charge tends to move toward lower $V$, because that lowers its potential energy. A negative charge tends to move toward higher $V$, because multiplying by a negative $q$ reverses the change in $U$.

Common mistakes

Confusing potential with electric field

The height of the graph gives $V$; its slope gives $E$. A large potential does not automatically mean a large electric field. A region may have a high but constant potential and therefore zero field.

Forgetting the negative sign

Using $E=\Delta V/\Delta x$ gives the wrong direction. The correct one-dimensional relation is

$$E_x=-\frac{\Delta V}{\Delta x}$$

for a constant field or a straight graph segment.

Mixing units

Convert centimetres to metres before calculating the slope. For example, a change of $12\ \mathrm{V}$ over $4\ \mathrm{cm}$ gives

$$\left|E\right|=\frac{12\ \mathrm{V}}{0.04\ \mathrm{m}}=300\ \mathrm{V/m},$$

not $3\ \mathrm{V/m}$.

Assuming the zero of potential is physical

Changing every value on the graph by the same constant changes the reference, not the electric field. The slope stays unchanged, so the physical electric field is unchanged.

A quick procedure

To analyze any position–potential graph:

1. Label the position and potential axes and check their units.
2. Identify the position where the field is requested.
3. Find the slope there, using a tangent for a curved graph.
4. Apply $E_x=-dV/dx$.
5. Report both the magnitude and the direction.
6. Check that the result has units of volts per metre or newtons per coulomb.