Electric field and potential difference

The electric field vector $\vec E$ at a point equals the force on a unit positive charge, while the potential $\varphi$ equals the potential energy of a unit positive charge at that point. The potential difference $U=\varphi_1-\varphi_2$ between two points equals the work done by the field when a unit positive charge is moved from point 1 to point 2. For a charge $q$, the work done by the field is $W=q(\varphi_1-\varphi_2)$ and is independent of the path, because the electrostatic field is conservative. In a uniform field the potential decreases linearly along $\vec E$, so $U=Ed,$ where $d$ is the displacement along a field line. In the general case the local relation is given by the derivative: $E=-\Delta\varphi/\Delta r$ (in the limit $E=-d\varphi/dr$), so $\vec E$ points toward the steepest decrease of potential and is perpendicular to equipotential surfaces.

A uniform electric field has magnitude $E=200\ \text{V/m}$ and points from plate A to plate B. The plates are $d=5.0\ \text{cm}$ apart. Find the potential difference $U=\varphi_A-\varphi_B$ and the work done by the field on a charge $q=+3.0\ \mu\text{C}$ moving from A to B.

A point charge $Q=+2.0\ \mu\text{C}$ is fixed in vacuum. Find the work done by its field on a charge $q=+1.0\ \text{nC}$ moving from $r_1=0.10\ \text{m}$ to $r_2=0.30\ \text{m}$ from $Q$. Use $k=9.0\times10^9\ \text{N·m}^2/\text{C}^2$.

In a region the potential is $\varphi(x,y)=300x-200y$ (volts with $x$ and $y$ in meters). (a) Find the vector $\vec E$. (b) Find the force on a charge $q=-2.0\ \mu\text{C}$. (c) Find the work done by the field when this charge moves from $(0,0)$ to $(1.0,0)$ m and from $(0,0)$ to $(0,1.0)$ m.

Safe self-study simulation plan: use an electric-field simulation (for example, PhET “Charges and Fields” or a similar one). Do not use mains high voltage. 1) Place two point charges and enable the grid and the potential probe. 2) Mark 5–6 points along the line between the charges and record the potential $\varphi$ at each point. 3) Compute the average $E\approx-\Delta\varphi/\Delta r$ over two neighboring intervals and compare the direction with the drawn field arrows. 4) Turn on equipotential lines and check that they are perpendicular to the arrows $\vec E$. 5) For a uniform field between parallel plates, measure $\varphi$ on two lines, find $U$, and check $U=Ed$. Record the results in a table and state the conclusion about the connection between field and potential difference.

Show that the potential difference equals the work done by the field per unit charge and that equipotential surfaces are perpendicular to field lines.

Sketch: two parallel plates, with evenly spaced arrows $\vec E$ from the positive plate to the negative plate; two dashed equipotential planes perpendicular to the arrows, labeled $\varphi_1$ and $\varphi_2$ ($\varphi_1>\varphi_2$); distance $d$ marked along a field line; formulas $U=\varphi_1-\varphi_2=Ed$ and $W=qU$ beside the diagram.

Before a lightning strike, the field strength in the air between the base of a thundercloud and the ground reaches about $E = 3\times10^{6}\ \text{V/m}$ (at this strength air breaks down). If the cloud sits at a height $d = 100\ \text{m}$ above the ground, the potential difference between cloud and ground is $U = E\,d = 3\times10^{6}\cdot100 = 3\times10^{8}\ \text{V}$. The discharge transfers about $q = 5\ \text{C}$ of charge, and the field does work $A = qU \approx 5\cdot3\times10^{8} = 1{,}5\times10^{9}\ \text{J}$ in a time of order $10^{-4}\ \text{s}$; the average power of the flash is $10^{13}\ \text{W}$, so most of the energy goes into heating the air and into thunder.

Both formulas of the theory are at work here: $E = U/d$ for a (approximately) uniform field converts field strength into voltage, and $U = A/q$ is the definition of potential difference as the field's work per unit charge transferred.

A power bank is labelled $3{,}7\ \text{V}$ and $3000\ \text{mA}\cdot\text{h}$. Convert the charge: $q = 3\ \text{A}\cdot3600\ \text{s} = 10800\ \text{C}$. The stored energy is then $A = qU = 10800\cdot3{,}7 \approx 4{,}0\times10^{4}\ \text{J}$, about $40\ \text{kJ}$ — roughly three full phone charges. For every coulomb transferred the battery delivers $3{,}7\ \text{J}$, while fast charging from a $5\ \text{V}$ supply gives $5\ \text{J}$ per coulomb.

The scenario exercises $U = A/q$ directly: voltage is the energy per unit charge, so capacity in $\text{C}$ multiplied by volts gives joules. Comparing $3{,}7\ \text{V}$ with $5\ \text{V}$ shows that a larger voltage means more work for the same charge.

A DC charging station supplies a voltage $U = 400\ \text{V}$ at a power $P = 100\ \text{kW}$. In half an hour ($t = 1800\ \text{s}$) the battery receives energy $A = P\,t = 1{,}0\times10^{5}\cdot1800 = 1{,}8\times10^{8}\ \text{J}$, that is $50\ \text{kW}\cdot\text{h}$. The charge passing through the cable is $q = A/U = 1{,}8\times10^{8}/400 = 4{,}5\times10^{5}\ \text{C}$, and each coulomb accounts for $400\ \text{J}$ of work done by the charging field.

It uses the definition of potential difference through work: $U = A/q$, hence $q = A/U$ and $A = qU$. Comparison with a phone battery ($3{,}7\ \text{V}$) shows that the same unit of transferred charge yields more energy the larger the voltage between the terminals.