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Electric Fields

Field strength calculations and combined fields

  • 3 things to know
  • Unit A2 2

Overview

Recall questions on the magnitude and direction of the field near a single negative charge, adding the fields of two like and two opposite charges at their midpoint and why the answers differ, balancing a charged oil drop against its weight, the acceleration of a proton in a uniform field, and accelerating a charge through a potential difference.

What you need to know

Answer each one before you reveal it.

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Numbers

Two OPPOSITE charges of magnitude 3.0 nC are 0.40 m apart. Calculate the field strength at the midpoint.

Numbers

A charge of −6.0 nC sits 0.15 m from a point. Calculate the magnitude of the field strength there.

Key term

A charge of −6.0 nC sits 0.15 m from a point. In which direction does the field at that point act?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    A charge of −6.0 nC sits 0.15 m from a point. Calculate the magnitude of the field strength there.

    Model answer

    E = 8.99 × 10⁹ × 6.0 × 10⁻⁹ ÷ 0.15², which is 54.0 ÷ 0.0225, giving 2.4 × 10³ N C⁻¹.

  2. 2

    In which direction does that field point?

    Model answer

    Towards the charge, because the charge is negative and the field direction is that of the force on a positive test charge.

  3. 3

    Two LIKE charges of 3.0 nC each are 0.40 m apart. Calculate the field strength at the midpoint.

    Model answer

    Each charge gives 8.99 × 10⁹ × 3.0 × 10⁻⁹ ÷ 0.20², which is 674 N C⁻¹, but the two point in opposite directions so the resultant is zero.

  4. 4

    Two OPPOSITE charges of magnitude 3.0 nC are 0.40 m apart. Calculate the field strength at the midpoint.

    Model answer

    Each still gives 674 N C⁻¹, but now both point the same way, from the positive towards the negative charge, so the resultant is 1.3 × 10³ N C⁻¹.

  5. 5

    An oil drop of weight 4.9 × 10⁻¹⁵ N is held stationary between plates giving a field of 3.0 × 10⁴ V m⁻¹. Calculate the charge on it.

    Model answer

    The electric force balances the weight, so q = F/E = 4.9 × 10⁻¹⁵ ÷ 3.0 × 10⁴, which is 1.6 × 10⁻¹⁹ C.

  6. 6

    Which way must the field point for the drop to be held up, given the drop is negatively charged?

    Model answer

    Downwards, so that the force on the negative charge is upwards and opposes the weight.

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