Electric Field between Parallel Plates
- The magnitude of the electric field strength in a uniform field between two charged parallel plates is defined as:
- Where:
- E = electric field strength (V m-1)
- V = potential difference between the plates (V)
- d = separation between the plates (m)
- The electric field strength is now defined by the units V m–1
- Therefore, the units V m–1 are equivalent to the units N C–1
- The equation shows:
- The greater the voltage (potential difference) between the plates, the stronger the field
- The greater the separation between the plates, the weaker the field
- Remember this equation cannot be used to find the electric field strength around a point charge (since this would be a radial field)
- The direction of the electric field is from the plate connected to the positive terminal of the cell to the plate connected to the negative terminal
The E field strength between two charged parallel plates is the ratio of the potential difference and separation of the plates
- Note: if one of the parallel plates is earthed, it has a voltage of 0 V
Worked example
Two parallel metal plates are separated by 3.5 cm and have a potential difference of 7.9 kV.
Calculate the electric force acting on a stationary charged particle between the plates that has a charge of 2.6 × 10-15 C.
Step 1: Write down the known values
- Potential difference, V = 7.9 kV = 7.9 × 103 V
- Distance between plates, d = 3.5 cm = 3.5 × 10-2 m
- Charge, Q = 2.6 × 10-15 C
Step 2: Calculate the electric field strength between the parallel plates
Step 3: Write out the equation for electric force on a charged particle
F = QE
Step 4: Substitute electric field strength and charge into electric force equation
F = QE = (2.6 × 10-15) × (2.257 × 105) = 5.87 × 10-10 N = 5.9 × 10-10 N (2 s.f.)
Examiner Tip
Remember the equation for electric field strength with V and d is only used for parallel plates, and not for point charges (where you would use E = F/Q)