OCR A Level Physics

Revision Notes

6.3.3 Electric Field Strength

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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:

Electric Field Strength Equation_2

  • 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

Force in an electric field

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 Worked Example

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:

E equals V over d

  • 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

10-1-1-electric-field-parallel-plates-ib-hl

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

Parallel Plates Work Done

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:

increment V equals W over Q

  • Therefore, the work done in transferring the charge is equal to:

W equals straight capital delta V cross times Q

  • When a charge Q moves from one plate to the other, its work done is:

W equals F cross times d

  • Where:
    • W = work done (J)
    • F = force (N)
    • d = distance (m)
  • Equate the expressions for work done:
F cross times d equals straight capital delta V cross times Q

  • Rearranging the fractions by dividing by Q and d on both sides gives:

F over Q equals fraction numerator straight capital delta V over denominator d end fraction

  • Since E equals F over Q the electric field strength between the plates can be written as:
E equals F over Q equals fraction numerator straight capital delta V over denominator d end fraction

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
E equals fraction numerator increment V over denominator d end fraction equals F over Q
Step 3: Rearrange for electric force, F

F equals fraction numerator Q increment V over denominator d end fraction

Step 4: Substitute the values into the electric force equation

F equals fraction numerator left parenthesis 2.6 cross times 10 to the power of negative 15 end exponent right parenthesis left parenthesis 7.9 cross times 10 cubed right parenthesis over denominator left parenthesis 3.5 cross times 10 to the power of negative 2 end exponent right parenthesis end fraction= 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 E equals F over Q)

Electric field strength parallel plates

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