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Electrostatic and Gravitational Fields of Force

Resultant fields and potentials from several sources

  • 3 things to know
  • Component 2

Overview

Recall questions on adding field strengths as vectors and potentials as scalars, why that difference matters, locating a point of zero resultant field, worked calculations for two charges or two masses, and why a zero field does not mean a zero potential.

What you need to know

Answer each one before you reveal it.

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Evaluation

How is the resultant field strength at a point due to several point charges found?

Key term

How is the resultant potential at a point due to several point charges found?

Key term

Why are fields added as vectors but potentials added as ordinary numbers?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How is the resultant field strength at a point due to several point charges found?

    Model answer

    The individual field strengths are calculated and then added as vectors, taking account of both their magnitudes and their directions.

  2. 2

    How is the resultant potential at a point due to several point charges found?

    Model answer

    The individual potentials are calculated and then added algebraically, taking account of their signs but not of any direction.

  3. 3

    Why are fields added as vectors but potentials added as ordinary numbers?

    Model answer

    Field strength is a force per unit charge and so has direction, while potential is a work done per unit charge and work is a scalar.

  4. 4

    Two equal positive point charges are held a fixed distance apart. What is the electric field strength at the midpoint of the line joining them?

    Model answer

    The field strength there is zero, because the two contributions are equal in size and point in opposite directions.

  5. 5

    Two point charges of +5.0 nC are held 0.40 m apart. Calculate the electric potential at the midpoint of the line joining them.

    Model answer

    Each charge contributes 9.0 × 10⁹ × 5.0 × 10⁻⁹ ÷ 0.20 = 225 V, so the total is 450 V. Dividing 450 V by two returns the 225 V from each charge.

  6. 6

    A charge of +5.0 nC and one of −5.0 nC are held 0.40 m apart. State the electric potential at the midpoint of the line joining them.

    Model answer

    The potential is zero, because the +225 V from one charge is exactly cancelled by the −225 V from the other.

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