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Force on a charge in an electric field

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The force relationship

Any charge placed in an electric field feels a force:

$F = qE$

$F$ — force on the charge (N)
$q$ — the size of the charge (C)
$E$ — electric field strength (N C⁻¹, or equivalently V m⁻¹)
The force is directly proportional to the charge: double the charge, double the force.
The force is directly proportional to the field strength: double the field, double the force.

Direction of the force

A positive charge is pushed in the direction of the field — toward the negative plate.
A negative charge is pushed against the field — toward the positive plate.
Because the field is uniform, the force has the same size and direction everywhere in the gap. The charge therefore undergoes constant acceleration, exactly like an object in free fall.

Combining the two relationships

Substituting $E = V/d$ into $F = qE$ gives a useful shortcut:

$F = \frac{qV}{d}$

Both routes are equally acceptable in an exam. Working in two steps — field first, then force — is safer, because it makes each substitution visible and each step separately markable.

Charge and the elementary charge

Charge is measured in coulombs (C).
The charge on a single proton is $+1.6 \times 10^{-19}$ C; on an electron, $-1.6 \times 10^{-19}$ C. This value is on the resource sheet.
A charge of $n$ electrons has size $q = n \times 1.6 \times 10^{-19}$ C.
Common prefixes:
- $1$ mC $= 10^{-3}$ C,
- $1$ µC $= 10^{-6}$ C,
- $1$ nC $= 10^{-9}$ C.

Comparing the electric force with weight

A charged particle between plates also has weight, $mg$ , acting downward.
For sub-atomic particles such as electrons and protons, the electric force is overwhelmingly larger than the weight — typically by more than ten orders of magnitude — so weight is ignored unless a question specifically asks about it.
For a larger object, such as a charged oil drop or a small sphere, the two can be comparable, and a question may ask you to balance them.

Worked ExampleForce on a charge between plates

Two parallel plates are $4.0$ mm apart with $200$ V across them. Find the force on a $3.0$ µC positive charge placed between them, and state its direction.

Step 1 — Convert units

$d = 4.0 \times 10^{-3} \text{ m}, \qquad q = 3.0 \times 10^{-6} \text{ C}$

Step 2 — Field strength

$E = \frac{V}{d} = \frac{200}{4.0 \times 10^{-3}} = 5.0 \times 10^{4} \text{ N C}^{-1}$

Step 3 — Force on the charge

$F = qE = (3.0 \times 10^{-6})(5.0 \times 10^{4}) = 0.15 \text{ N}$

Step 4 — Direction

The charge is positive, so the force is in the direction of the field — toward the negative plate.

$F = 0.15 \text{ N toward the negative plate}$

Worked ExampleBalancing an electric force against weight

A small charged sphere of mass $2.0 \times 10^{-6}$ kg hangs motionless between two horizontal parallel plates $5.0$ mm apart. The plates have $490$ V across them. Find the charge on the sphere.

Step 1 — Recognise what "motionless" means

The sphere is in equilibrium, so the upward electric force exactly balances its downward weight:

$F_{\text{electric}} = F_{\text{weight}}$ $qE = mg$

Step 2 — Find the field strength

$E = \frac{V}{d} = \frac{490}{5.0 \times 10^{-3}} = 9.8 \times 10^{4} \text{ N C}^{-1}$

Step 3 — Find the weight

$mg = (2.0 \times 10^{-6})(9.8) = 1.96 \times 10^{-5} \text{ N}$

Step 4 — Solve for the charge

$q = \frac{mg}{E} = \frac{1.96 \times 10^{-5}}{9.8 \times 10^{4}} = 2.0 \times 10^{-10} \text{ C}$

$q = 2.0 \times 10^{-10} \text{ C} = 0.20 \text{ nC}$