Electric field of a point charge
- The strength of an electric field due to a point charge decreases with the square of the distance
- This is an inverse square law, similar to Coulomb's law
- Using Coulomb's law and the electric field strength equation this can be written as:
- Where:
- Q = the point charge producing the radial electric field (C)
- r = distance from the centre of the charge (m)
- ε0 = permittivity of free space (F m−1)
- Here, q represents the magnitude of a small positive test charge
- This equation shows that in a radial electric field, the electric field strength E:
- is not constant
- follows an inverse square law with distance r
- This equation is analogous to the gravitational field strength around a point mass
Graph of field strength against distance for a positive charge
Electric field strength is zero inside a charged sphere and decreases with distance outside the sphere according to an inverse square law
Combining electric fields
- Both electric force and field strength are vector quantities
- If the charge is negative, the E field strength is negative and points towards the centre of the charge
- If the charge is positive, the E field strength is positive and points away from the centre of the charge
- Therefore, to find the electric force or field strength at a point due to multiple charges, each field can be combined by vector addition
Vector addition of electric field along the same line
For charges along the same line, the resultant field is the vector addition of the field due to both charges at a particular point
- For a point on the same line as two charges q1 and q2, with field strengths E1 and E2 respectively, the magnitude of the resultant field will be:
- The sum of the fields, E1 + E2, if they are both in the same direction
- The difference between the fields, E1 − E2, if they are in opposite directions
- The direction of the resultant field depends on:
- The types of charge (positive or negative)
- The magnitude of the charges
- For a point which makes a right-angled triangle with the charges, the resultant field can be determined using Pythagoras theorem
Vector addition of electric field components
For charges which make a right-angle triangle with point X, the resultant field is the vector addition of the field due to both charges using Pythagoras theorem
Worked example
A metal sphere of diameter 15 cm is negatively charged. The electric field strength at the surface of the sphere is 1.5 × 105 V m−1.
Determine the total surface charge of the sphere.
Answer:
Step 1: List the known quantities
- Electric field strength, E = 1.5 × 105 V m−1
- Radius of sphere, r = 15 / 2 = 7.5 cm = 7.5 × 10−2 m
- Permittivity of free space, ε0 = 8.85 × 10−12 F m−1
Step 2: Write down the equation for electric field strength
Step 3: Rearrange for charge Q
- Recall that, for an external observer, a charged sphere can be considered as a point charge
- The field strength at the surface of the sphere is equal to the field strength 15 cm away from a point charge with the same total charge as the sphere
Step 4: Substitute in the values and calculate:
Q = (4π × 8.85 × 10−12) × (1.5 × 105) × (7.5 × 10−2)2 = 9.38 × 10−8 C
- The particle has a charge of 9.4 × 10−8 C or 94 nC
Examiner Tip
When combining electric fields from multiple charges, remember that the point (e.g. point X in the examples above) represents a positive test charge, so the direction of the electric force or field will correspond to the signs of the charges; the direction of the force or field points away from a positive charge and towards a negative charge.