To calculate the magnitude of the potential difference between C and D:

List the key information:

Calculate the potential at C due to X and Y:

$V_{C X} = \frac{5.0 \times 10^{- 7}}{4 π \times (8.85 \times 10^{- 12}) \times 0.025} = 1.8 \times 10^{5} V$
$V_{C Y} = \frac{- 5.0 \times 10^{- 7}}{4 π \times (8.85 \times 10^{- 12}) \times 0.055} = - 0.8 \times 10^{5} V$
Maximum of one mark for either calculation

Calculate the potential at D due to X and Y:

$V_{D X} = \frac{5.0 \times 10^{- 7}}{4 π \times (8.85 \times 10^{- 12}) \times 0.055} = 0.8 \times 10^{5} V$
$V_{D Y} = \frac{- 5.0 \times 10^{- 7}}{4 π \times (8.85 \times 10^{- 12}) \times 0.025} = - 1.8 \times 10^{5} V$
Maximum of one mark for either calculation

Calculate total potentials at C and D:

Calculate p.d. at the midpoint:

[Total: 5 marks]

Electric Fields (Edexcel International A Level Physics)

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