Define the electric field as force per unit charge, $E = F_e/q$, describe the uniform field between parallel plates with $E = V/d$, and define electric potential difference as work per unit charge, $V = W/q$.

What this topic is asking

This dot point moves from the force between charges to the field that mediates it and the potential difference that drives charge through circuits. The Physical Setting/Physics course asks you to define the electric field as force per unit charge, $E = \dfrac{F_e}{q}$, to describe the uniform field between parallel plates with $E = \dfrac{V}{d}$, and to define electric potential difference (voltage) as work per unit charge, $V = \dfrac{W}{q}$. The Regents tests these as calculations and as field-line interpretation.

The electric field

The field idea lets us describe the influence of charges on the space around them without referring to a second charge until one is placed there. Once the field $E$ at a point is known, the force on any charge $q$ placed there is $F_e = qE$. The field strength is a property of the source charges and the location, not of the test charge: using a larger test charge gives a proportionally larger force, so the ratio $F_e/q$ is unchanged.

Field-line diagrams

Reading and sketching field lines is a common Regents task. The density of the lines indicates the field strength, and their direction gives the force direction on a positive charge (the opposite for a negative charge). The uniform field between parallel plates is the most important case for calculation.

The uniform field between parallel plates

Between two parallel plates connected to a source, the field is uniform and given by

where $V$ is the potential difference across the plates and $d$ is their separation. Because the field is uniform, the force on a charge is the same everywhere between the plates, $F_e = qE$. The units N/C and V/m are equivalent, which this equation makes clear. This setup is the basis of capacitors and of devices that accelerate charges.

Electric potential difference

Potential difference is what drives charge around a circuit, the "push" supplied by a battery. Moving a charge $q$ through a potential difference $V$ does work $W = qV$ on it, which is how energy is delivered to circuit components. This links directly to the circuit equations in current and Ohm's law, where the same voltage drives a current through a resistance.

Reference Tables note

The Reference Tables print $E = \dfrac{F_e}{q}$, $E = \dfrac{V}{d}$ and $V = \dfrac{W}{q}$ in the Electricity section. The field-line conventions are a stated idea you recall, and $F_e = qE$ is the rearrangement of the printed field definition. Potential difference is the same $V$ that appears in Ohm's law and the power equations for circuits.

Try this

Q1. A $3.0 \times 10^{-6}$ C charge feels a force of $0.012$ N in a field. Calculate the field strength. [2 points]

• Cue. $E = \dfrac{F_e}{q} = \dfrac{0.012}{3.0 \times 10^{-6}} = 4.0 \times 10^3$ N/C.

Q2. State the direction of the electric field around a single negative charge. [1 point]

• Cue. Radially inward, toward the charge (the direction a positive test charge would be pushed).