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The diagram shows a uniform electric field of strength 15 V m⁻¹ - AQA - A-Level Physics - Question 16 - 2022 - Paper 2

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The diagram shows a uniform electric field of strength 15 V m⁻¹. The length RS is perpendicular to the field and the line ST is parallel to the field. What is the t... show full transcript

Worked Solution & Example Answer:The diagram shows a uniform electric field of strength 15 V m⁻¹ - AQA - A-Level Physics - Question 16 - 2022 - Paper 2

Step 1

Calculate the Electric Potential Difference

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Answer

To find the electric potential difference (V), we can use the formula:

V=EimesdV = E imes d

where:

  • E is the electric field strength (15 V/m)
  • d is the distance moved in the direction of the field (3.0 m)

Thus,

V=15extV/mimes3.0extm=45extVV = 15 ext{ V/m} imes 3.0 ext{ m} = 45 ext{ V}

Step 2

Calculate the change in electrical potential energy

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Answer

The change in electrical potential energy (ΔU) can then be calculated using:

ΔU=qimesVΔU = q imes V

where:

  • q is the charge (3.0 µC = 3.0 \times 10^{-6} C)
  • V is the potential difference (45 V)

Thus,

ΔU=3.0×10−6extC×45extV=135×10−6extJ=135extµJΔU = 3.0 \times 10^{-6} ext{ C} \times 45 ext{ V} = 135 \times 10^{-6} ext{ J} = 135 ext{ µJ}

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