The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
Learn more about potential difference across parallel plate capacitor here:
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Answer:
Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.
Answer:
<h3>The answer is 45 J</h3>
Explanation:
The work done by an object can be found by using the formula
<h3>workdone = force × distance</h3>
From the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
<h3>45 J</h3>
Hope this helps you
