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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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Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

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slava [35]

Answer:

Hey I would say Dust

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2 years ago
Boiling water is a common method used to sterilize water that may be contaminated. subjecting microbes to such high temperatures
GaryK [48]

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8 0
2 years ago
A 500-kilogram sports car accelerates uni-
timama [110]

Answer:

90 meters

Explanation:

Given:

x₀ = 0 m

v₀ = 0 m/s

v = 30 m/s

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x

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