# Electric Field Strength(OCR A Level Physics)

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Katie M

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Physics

## 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 C1 where it experiences a force of 0.3 N.

Calculate the charge of the particle.

## 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 m1
• 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.

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 × 1015 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

#### Exam 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 )

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