1.1.10Electricity & Charge Basics

Define electric field and electric potential

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WHY do we even need these ideas?

If charge A and charge B interact, we could just write the force between them (Coulomb's law). But that couples the two charges together. Physicists wanted to say: "charge A alone prepares the space; then whatever charge you drop in, you just read off the preparation."

  • The field answers: "If I put a +1 C test charge here, how hard and which way is it shoved?"
  • The potential answers: "How much work per coulomb did I do to get here from infinity?"

This separation is the source vs. test charge idea, and it is the foundation of all circuit theory (voltage = potential difference).


Electric Field

WHY divide by q0q_0? Because we want a property of the space, not of the visitor. Doubling the test charge doubles the force, so F/q0\vec F/q_0 stays constant — it describes the location itself.

Deriving the field of a point charge (from scratch)

Start from Coulomb's law — the measured force between two point charges: F=14πε0Qq0r2r^\vec{F} = \frac{1}{4\pi\varepsilon_0}\frac{Qq_0}{r^2}\hat{r}

Now apply the definition E=F/q0\vec E = \vec F / q_0 — divide out the test charge: E=14πε0Qr2r^\boxed{\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}}


Electric Potential

Deriving potential of a point charge (from scratch)

Work is force times distance. To bring q0q_0 from infinity to rr, we push against the field, so: W=rFelecdl=rq0Edr=rq0kQr2drW = -\int_\infty^r \vec{F}_{elec}\cdot d\vec{l} = -\int_\infty^r q_0 E\,dr = -\int_\infty^r q_0\frac{kQ}{r^2}\,dr

Why the minus sign? The applied work equals minus the work done by the electric force (we oppose it).

Integrate (r2dr=r1\int r^{-2}dr = -r^{-1}): W=q0kQ[1r]r=q0kQ(1r1)=kQq0rW = -q_0 kQ\left[-\frac{1}{r}\right]_\infty^r = q_0 kQ\left(\frac{1}{r}-\frac{1}{\infty}\right)=\frac{kQq_0}{r}

Divide by q0q_0 (definition of VV): V=kQr\boxed{V = \frac{kQ}{r}}

E=dVdrV(r)=rEdrE = -\frac{dV}{dr} \qquad\Longleftrightarrow\qquad V(r) = -\int_\infty^r E\,dr

Figure — Define electric field and electric potential

Worked Examples


Common Mistakes


Active Recall

Recall Test yourself (hide the answers)
  • What are the units of EE? → N/C=V/m\text{N/C}=\text{V/m}.
  • What are the units of VV? → J/C=V\text{J/C}=\text{V}.
  • Is EE scalar or vector? Is VV? → EE vector, VV scalar.
  • Why does q0q_0 cancel in EE and VV? → So they describe the space, not the visiting charge.
  • Relationship between EE and VV? → E=dV/drE=-dV/dr; field is the negative slope of potential.
Recall Feynman: explain to a 12-year-old

Imagine a hill made by a magnet-like ball. The potential is how high you are on the hill — being high up means you have stored energy. The field is how steep the hill is right under your feet — steep means a big push. A ball always rolls from a high spot to a low spot, and it rolls fastest where the hill is steepest. Voltage in your phone charger is just "how high the electricity hill is," and that's what pushes the tiny charges through the wire.


Connections

Electric field definition
Force per unit positive test charge, E=F/q0\vec E=\vec F/q_0; a vector in N/C.
Electric potential definition
Work done per unit charge to bring a positive charge from infinity, V=W/q0V=W/q_0; a scalar in volts (J/C).
Field of a point charge
E=kQ/r2E=kQ/r^2, direction away from + / toward −.
Potential of a point charge
V=kQ/rV=kQ/r (scalar, no direction).
Why does field fall as 1/r² but potential as 1/r?
Potential is the integral of field over distance; r2drr1\int r^{-2}dr\propto r^{-1}.
Relation between E and V
E=dV/drE=-dV/dr; field is the negative slope (gradient) of potential.
Units of E and V
EE: N/C = V/m; VV: J/C = volt.
Can E=0 while V≠0?
Yes — e.g. inside a hollow charged conductor.
Can V=0 while E≠0?
Yes — e.g. midpoint between equal + and − charges.
Force on charge q in field E
F=qEF=qE; direction reverses for negative q.
Work by field moving q between potentials
W=q(ViVf)W=q(V_i-V_f).
Value of Coulomb constant k
k=1/(4πε0)8.99×109 N⋅m2/C2k=1/(4\pi\varepsilon_0)\approx 8.99\times10^9\ \text{N·m}^2/\text{C}^2.

Concept Map

creates

divide by q0

defines

has direction

integrate force over path

defines

just a number

difference gives

independent of

Source charge Q prepares space

Coulomb's law force

Field = force per unit charge

Electric field vector E

Potential = work per unit charge

Electric potential scalar V

Test charge q0

Voltage = potential difference

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi charge QQ akela space mein rakha hota hai, to woh apne aas-paas ki jagah ko "influence" kar deta hai. Is influence ko hum do tareeke se samajhte hain. Pehla hai electric field — matlab agar us jagah par ek chhota positive test charge rakhein to uspe kitni force lagegi, per unit charge. Formula: E=F/q0\vec E=\vec F/q_0, units N/C, aur ye ek vector hai (direction hoti hai). Field point charge ke liye E=kQ/r2E=kQ/r^2 hota hai.

Doosra hai electric potential — matlab us test charge ko infinity se us point tak laane mein kitna kaam (work) karna pada, per unit charge. Formula V=W/q0V=W/q_0, units volt (J/C), aur ye ek scalar hai (sirf number, koi direction nahi). Point charge ke liye V=kQ/rV=kQ/r. Dhyan do: field 1/r21/r^2 se girta hai lekin potential 1/r1/r se — kyunki potential asal mein field ka integral over distance hai.

Sabse important intuition: potential ek pahaadi ki height jaisa hai, aur field us pahaadi ka dhalaan (slope) hai — E=dV/drE=-dV/dr. Charge hamesha high potential se low potential ki taraf "roll" karta hai, aur wahan sabse fast jaata hai jahan slope sabse steep ho. Yehi cheez circuit mein "voltage" ban jaati hai jo current ko push karti hai.

Common galti: log sochte hain jahan field strong hai wahan potential bhi high hoga — galat! Field to slope hai, value nahi. Do plates ke beech field constant rehta hai par potential lagataar badalta hai. Yaad rakho: "Vector Force, Scalar Store" — E force wala vector, V energy wala scalar.

Go deeper — visual, from zero

Test yourself — Electricity & Charge Basics

Connections