Potential & Potential Energy

PHYPotential & Capacitance

Electric potential $V$ at a point is the work done per unit positive charge to bring it from infinity to that point, a scalar quantity measured in volts. Potential energy $U$ of a charge configuration is the work done in assembling the charges from infinity.

Because potential is a scalar, problems with several charges are much easier than the corresponding force problems.

Potential of a point charge

V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}

V

in volts; carries the sign of

q

(negative for a negative charge), zero at

r\to\infty

Derivation

Potential is the work per unit charge to bring a test charge from infinity to the point — the line integral of the field:
$V = -\int_{\infty}^{r} \vec{E}\cdot d\vec{l} = -\int_{\infty}^{r} \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r'^2}\,dr'$
Integrating $1/r'^2$ gives $-1/r'$ , evaluated from $\infty$ to $r$ :
$V = -\dfrac{q}{4\pi\varepsilon_0}\left[-\dfrac{1}{r'}\right]_{\infty}^{r} = \dfrac{q}{4\pi\varepsilon_0}\left(\dfrac{1}{r}-0\right)$
Hence the potential of a point charge is:
$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$

Potential due to several charges

V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}

algebraic (scalar) sum of individual potentials with their signs;

r_i

is distance from each charge to the point

Derivation

Potential is a scalar, so the net value is the ordinary (algebraic) sum of what each charge produces alone:
$V = V_1 + V_2 + \cdots + V_n$
Substituting the point-charge potential $V_i = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i}{r_i}$ for each charge:
$V = \dfrac{1}{4\pi\varepsilon_0}\sum_i \dfrac{q_i}{r_i}$
Each $q_i$ carries its own sign; no vector resolution is needed because potentials add as numbers.

PE of a pair of charges

U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r}

U>0

for like charges (repulsive),

U<0

for unlike charges (attractive)

Derivation

Bring $q_1$ in first: no field exists yet, so this costs no work. Now bring $q_2$ from infinity to distance $r$ against the field of $q_1$ . The work equals $q_2$ times the potential of $q_1$ there:
$U = q_2 V_1 = q_2\left(\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r}\right)$
This stored work is the potential energy of the pair:
$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r}$

PE of a system of charges

U = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_2}{r_{12}} + \dfrac{q_2 q_3}{r_{23}} + \dfrac{q_1 q_3}{r_{13}}\right)

sum over every distinct pair once; each

r_{ij}

is the separation of that pair

Derivation

Assemble the charges one at a time; each new charge does work against those already in place. The total is the sum over every distinct pair:
$U = U_{12} + U_{23} + U_{13}$
Writing each pair energy as $U_{ij} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_i q_j}{r_{ij}}$ :
$U = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1 q_2}{r_{12}} + \dfrac{q_2 q_3}{r_{23}} + \dfrac{q_1 q_3}{r_{13}}\right)$
Each pair is counted once, so a system of $n$ charges has $\dfrac{n(n-1)}{2}$ terms.

Work done moving a charge

W = q\,(V_B - V_A)

W

is work by an external agent to move charge

q

from A to B; depends only on the endpoint potentials

Derivation

Potential difference is defined as the work done per unit charge in moving between two points:
$V_B - V_A = \dfrac{W}{q}$
Rearranging gives the work an external agent does (moving $q$ slowly) from A to B:
$W = q\,(V_B - V_A)$
The electrostatic field is conservative, so $W$ depends only on the endpoints, not the path taken.

Potential is a scalar — add the contributions of different charges algebraically (with signs), never as vectors.
Work done by the electrostatic force as a charge moves equals $-\Delta U$ ; the field is conservative, so work is path-independent.
Potential at a point can be zero while the field there is non-zero (e.g. the midpoint between equal and opposite charges).
For an equilateral triangle of equal charges each pair has the same separation, so $U = 3\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{a}$ .
$1\,\text{V}=1\,\text{J C}^{-1}$ , and the relation to field is $E=-\dfrac{dV}{dr}$ (field points from high to low potential).
When moving a charge between two points, only the potential difference matters, not the path or the field shape between them.
Keep track of signs of charges throughout — a negative charge lowers the potential and an unlike pair has negative potential energy.

Where the marks go

Adding potentials as vectors or ignoring the sign of negative charges — potential is signed and scalar.
Counting each pair twice (or missing a pair) in the system PE; three charges give exactly $3$ pairs.
Mixing up $V$ (potential, scalar, volts) and $E$ (field, vector, $\text{N C}^{-1}$ ); they have different units and behaviours.
Using $W=qV$ instead of $W=q(V_B-V_A)$ — only the difference in potential does work.

How the board asks it

Numericalpotential due to several charges added algebraically
Two point charges $q_1 = +5\,\mu C$ and $q_2 = -3\,\mu C$ are placed $20\,cm$ apart. Calculate the electric potential at the midpoint of the line joining them.
NumericalPE of a system of charges
Three charges of $+2\,\mu C$ each are placed at the vertices of an equilateral triangle of side $10\,cm$ . Calculate the total electrostatic potential energy of the system.
Numericalwork done equals charge times potential difference
A charge of $+4\,\mu C$ is moved from a point $A$ at potential $30\,V$ to a point $B$ at potential $10\,V$ . Calculate the work done by the electrostatic force in moving the charge.
Give reasonspotential can be zero while field is non-zero
At a point on the line joining two equal and opposite point charges, the electric potential is zero but the electric field is not zero. Explain why this is so.
Define / stateelectric potential and the volt
Define electric potential at a point and state its SI unit. State whether it is a scalar or a vector quantity.
Assertion–Reasonwork is path-independent in a conservative field
Assertion: The work done in moving a charge between two points in an electrostatic field is independent of the path followed. Reason: The electrostatic field is conservative. Select the correct option regarding these two statements.

Practise this topic

PreviousElectric Flux & Gauss's Law NextCapacitance, Dielectrics & Energy

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.