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Electric and Magnetic Fields

Uniform fields between parallel plates, and the potential near a point charge

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  • Papers 1 and 3

Overview

Recall questions on E = V/d, the field between charged plates and where it stops being uniform, and the potential a distance r from a point charge.

What you need to know

Answer each one before you reveal it.

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What is the potential a distance r from a point charge, how does it vary with distance, and what happens to the field strength between parallel plates if their separation is halved at constant potential difference?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Two parallel plates 5.0 mm apart have a potential difference of 250 V across them. What is the field strength between them?

    Model answer

    E = V/d = 250 ÷ (5.0 × 10⁻³) = 5.0 × 10⁴ V m⁻¹.

  2. 2

    A charge of 2.0 nC is placed in a uniform field of strength 5.0 × 10⁴ V m⁻¹. What force acts on it?

    Model answer

    F = EQ = 5.0 × 10⁴ × 2.0 × 10⁻⁹ = 1.0 × 10⁻⁴ N.

  3. 3

    What happens to the field strength between parallel plates if the plate separation is halved at constant potential difference?

    Model answer

    The field strength doubles, because E is inversely proportional to d.

  4. 4

    State the equation for the potential at a distance r from a point charge:

    Model answer

    V = Q/(4πε₀r), where Q is the point charge.

  5. 5

    How does the potential due to a point charge vary with distance?

    Model answer

    It is inversely proportional to the distance, falling as 1/r and tending to zero at infinity.

  6. 6

    What is the sign of the potential at a point near an isolated negative charge?

    Model answer

    It is negative, because work must be done against the attraction to remove a positive charge to infinity.

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