Electric Field Strength (OCR A Level Physics)
Revision Note
Electric Field Strength
An electric field is a region of space in which an electric charge “feels” a force
The electric field strength at a point is defined as:
The electrostatic force per unit positive charge acting on the charge at that point
The electric field strength can be calculated using the equation:
Where:
E = electric field strength (N C−1)
F = electrostatic force on the charge (N)
Q = charge (C)
It is important to use a positive test charge in this definition, as this determines the direction of the electric field
The electric field strength is a vector quantity, it is always directed:
Away from a positive charge
Towards a negative charge
Worked Example
A charged particle is in an electric field with electric field strength 3.5 × 104 N C−1 where it experiences a force of 0.3 N.
Calculate the charge of the particle.
Answer:
Electric Field Strength in a Uniform Field
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)
Note: the electric field strength is now also defined by the units V m−1
The equation shows:
The greater the voltage between the plates, the stronger the field
The greater the separation between the plates, the weaker the field
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
Derivation of Electric Field Strength Between Plates
When two points in an electric field have a different potential, there is a potential difference between them
To move a charge across that potential difference, work needs to be done
Two parallel plates with a potential difference ΔV across them create a uniform electric field
The work done on the charge depends on the electric force and the distance between the plates
Potential difference is defined as the energy, W, transferred per unit charge, Q, this can also be written as:
Therefore, the work done in transferring the charge is equal to:
When a charge Q moves from one plate to the other, its work done is:
Where:
W = work done (J)
F = force (N)
d = distance (m)
Equate the expressions for work done:
Rearranging the fractions by dividing by Q and d on both sides gives:
Since the electric field strength between the plates can be written as:
Worked Example
Two parallel metal plates are separated by 3.5 cm and have a potential difference of 7.9 kV.
Calculate the magnitude of the electric force acting on a stationary charged particle between the plates that has a charge of 2.6 × 10-15 C.
Answer:
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: Write down the equation for the electric field strength between the parallel plates
Step 3: Rearrange for electric force, F
Step 4: Substitute the values into the electric force equation
= 5.869 × 10−10 N
Step 5: State the final answer
The magnitude of the electric force acting on this charged particle is 5.9 × 10−10 N
Examiner Tips and Tricks
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 )
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