EG1216 · Physics 2 — Electricity & Magnetism
Theme 2 · Electric potential & capacitance

Potential & PE of point charges; E from V

Find the actual voltage around a charge, add voltages as plain numbers, store energy in a pair of charges, and turn voltage back into field.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §25.3–25.4.

Before you start

What you need first

  • Topic 9 — potential \(V\) as energy per charge, and \(U = qV\).
  • Topic 2 — Coulomb's constant \(k = 8.99\times10^{9}\).
  • Topic 3 — the field of a point charge.

What you'll be able to do

  • Find the potential around a point charge, \(V = kQ/r\) (with sign).
  • Add the potentials of several charges — just numbers.
  • Find the energy of a pair of charges, \(U = kq_1q_2/r\).
  • Get the field back from how the potential changes, \(E = -\Delta V/\Delta x\).

The potential around a charge

The potential of a point charge

A single charge \(Q\) sets up a potential everywhere around it. Like the field, it gets weaker with distance — but as \(1/r\), not \(1/r^2\), and it is just a number:

$$ V = \frac{kQ}{r} $$

Keep the sign of \(Q\): a \(+\) charge makes a positive potential (a "hill"), a \(-\) charge a negative one (a "valley").

Q
Dashed circles are equipotentials (same \(V\)); the red field lines cross them at right angles.
📐 Worked example 1

Voltage near a charge

(a) Find the potential \(0.30\) m from \(q = +5.0\ \mu\text{C}\). (b) Find it \(0.10\) m from \(q = -2.0\ \mu\text{C}\).

1Positive charge, \(V = kQ/r\):
$$ V = \frac{(8.99\times10^{9})(5.0\times10^{-6})}{0.30} \approx +1.5\times10^{5}\ \text{V} $$
2Negative charge — the sign carries through:
$$ V = \frac{(8.99\times10^{9})(-2.0\times10^{-6})}{0.10} \approx -1.8\times10^{5}\ \text{V} $$
A negative charge makes a negative potential around it.

Several charges? Just add the potentials

Because potential is a scalar, the total at a point is each charge's potential added up — with signs, but no arrows and no angles:

$$ V = k\sum \frac{q_i}{r_i} = k\!\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots\right) $$
This is the payoff of voltage over field: adding fields means adding vectors (components, angles); adding potentials means adding ordinary signed numbers.
📐 Worked example 2

Adding two voltages

At a point \(P\), charge \(q_1 = +4.0\ \mu\text{C}\) is \(0.20\) m away and \(q_2 = -3.0\ \mu\text{C}\) is \(0.30\) m away. Find the total potential at \(P\).

1Add each charge's potential (with signs):
$$ V = (8.99\times10^{9})\!\left(\frac{4.0\times10^{-6}}{0.20} + \frac{-3.0\times10^{-6}}{0.30}\right) $$
2The brackets give \(2.0\times10^{-5} - 1.0\times10^{-5} = 1.0\times10^{-5}\):
$$ V \approx +9.0\times10^{4}\ \text{V} $$

The energy stored in a pair of charges

Two charges a distance \(r\) apart store electric potential energy. (It follows from \(U = qV\): charge \(q_2\) sits at the potential \(kq_1/r\) made by \(q_1\).)

$$ U = \frac{kq_1q_2}{r} $$
where:
SymbolMeaningSI unit
Uenergy stored in the pairJ
q₁, q₂the two charges (keep their signs!)C
rdistance between them (note: \(r\), not \(r^2\))m
Like charges give \(U>0\) (energy stored, like a squeezed spring, they push apart). Opposite charges give \(U<0\) (stuck together; energy is needed to pull them apart).
📐 Worked example 3

Energy of two charges

(a) Like charges: \(q_1 = +3.0\ \mu\text{C}\), \(q_2 = +2.0\ \mu\text{C}\), \(r = 0.20\) m. (b) Opposite: \(q_1 = +4.0\ \mu\text{C}\), \(q_2 = -2.0\ \mu\text{C}\), \(r = 0.10\) m.

1Like charges, \(U = kq_1q_2/r\):
$$ U = \frac{(8.99\times10^{9})(3.0\times10^{-6})(2.0\times10^{-6})}{0.20} \approx +0.27\ \text{J} $$
2Opposite charges — the sign carries through:
$$ U = \frac{(8.99\times10^{9})(4.0\times10^{-6})(-2.0\times10^{-6})}{0.10} \approx -0.72\ \text{J} $$

Turning it around

Getting the field back from the potential

The field points "downhill" in potential, and its strength is how fast the potential drops with distance:

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

The minus sign says \(E\) points toward lower \(V\). Steep drop → strong field; flat \(V\) → no field. For a uniform field this is just Topic 9's \(E = V/d\).

r V V = kQ/r slope → E
\(V\) falls as \(1/r\). The steeper the slope, the stronger the field at that point.

✏️ Try it yourself

Take a charge \(q_1 = +6.0\ \mu\text{C}\).

(a) Find the potential \(0.40\) m from it.
(b) A second charge \(q_2 = -2.0\ \mu\text{C}\) is placed \(0.40\) m from \(q_1\). Find the potential energy of the pair.

(a) \(V = \dfrac{kQ}{r} = \dfrac{(8.99\times10^{9})(6.0\times10^{-6})}{0.40} \approx +1.3\times10^{5}\ \text{V}.\) (b) \(U = \dfrac{kq_1q_2}{r} = \dfrac{(8.99\times10^{9})(6.0\times10^{-6})(-2.0\times10^{-6})}{0.40} \approx -0.27\ \text{J}\) (negative — opposite charges).

Common mistakes

MistakeFix
Using \(r^2\) in \(V\) or \(U\).Potential and PE go as \(1/r\) (only the field and force use \(1/r^2\)).
Dropping the signs of the charges.For \(V\) and \(U\), keep \(+/-\) — they decide the sign of the answer.
Adding potentials as vectors.\(V\) is a scalar; add the signed numbers, no components.
Forgetting \(E\) points toward lower \(V\).\(E = -\Delta V/\Delta x\): downhill in potential.

Recap — the whole topic on one screen

IdeaWhat you own now
Point charge\(V = \dfrac{kQ}{r}\) (keep the sign); falls as \(1/r\).
Superposition\(V = k\sum q_i/r_i\) — add the signed numbers.
Pair of charges\(U = \dfrac{kq_1q_2}{r}\); like → \(+\), opposite → \(-\).
Field from V\(E = -\dfrac{\Delta V}{\Delta x}\) — points toward lower \(V\).

Next topic

Potential of distributions & conductors

Next we look at the potential of spread-out charge and of a metal: a charged conductor is all at one voltage, and its surface is an equipotential — which is why field lines always meet it at right angles.

→ Topic 11 · Potential of distributions & conductors