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Two parallel metal plates separated by a distance d have a potential difference V across them - AQA - A-Level Physics - Question 16 - 2021 - Paper 2

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Two parallel metal plates separated by a distance d have a potential difference V across them. A particle with charge Q is placed midway between the plates. What is... show full transcript

Worked Solution & Example Answer:Two parallel metal plates separated by a distance d have a potential difference V across them - AQA - A-Level Physics - Question 16 - 2021 - Paper 2

Step 1

What is the magnitude of the electrostatic force acting on the particle?

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Answer

To find the electrostatic force acting on a charge in an electric field, we can use the formula:

F=QEF = QE

where:

  • FF is the force,
  • QQ is the charge,
  • EE is the electric field strength.
  1. Calculate the Electric Field Strength (E): For two parallel plates, the electric field is given by: E=VdE = \frac{V}{d} where:

    • VV is the potential difference,
    • dd is the separation between the plates.
  2. Substitute E in the Force Equation: Substituting for EE: F=Q(Vd)F = Q \left(\frac{V}{d}\right) Thus, the expression for the force becomes: F=QVdF = \frac{QV}{d}

This means the magnitude of the electrostatic force acting on the particle is: rac{QV}{d}.

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