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Nikolay [14]
3 years ago
13

Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential differen

ce is applied across the plates? (Choose from: Increasing the area of the plates; Decreasing the area of the plates; Increasing the separation between the plates; Decreasing the separation between the plates; Inserting a material with a higher dielectric constant between the plates
Physics
1 answer:
Debora [2.8K]3 years ago
6 0

Answer:

increasing the separation between the plates

Explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV  V = VOLTAGE , c = capacitance  

E = 1/2 eAV^2/ D  ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

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A 100 N force is applied to a 500 kg crate resting on frictionless wheels.
Alla [95]

Answer:

<h2>0.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{100}{500}  =  \frac{1}{5}  = 0.2 \\

We have the final answer as

<h3>0.2 m/s²</h3>

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Answer:

None

Explanation:

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Elementary particles are the particles without any sub-structure which means they are not composed of other particles.

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