Respuesta :
Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:
[tex]C=\frac{\epsilon_0.A}{d}[/tex]
[tex]\therefore C\propto \frac{1}{d}[/tex]
In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where:
charge density, [tex]\sigma=\frac{Q}{A}[/tex]
As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.
iv)
- From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.
- Here we increase the distance so the work done per unit charge increases.
i) Capacitance is decreases
ii) Charge Q remains constant
iii) Electric field E remains constant
iv) Change in potential ΔV is increases
Parallel-plate capacitor:
The capacitance is computed as,
[tex]C=\frac{\epsilon A}{d}[/tex]
Where A is area of plates and d is distance between plates.
Following information is to be considered.
- Given that the distance between the plates increases, the capacitance decreases while area and permittivity of free space remains constant.
- the amount of charge is independent on plate separation .Hence the charge Q remains constant.
- We know that distance of plate separation can not affect area of the plate. So that Charge Q and permittivity are also not affected by it. Thus, electric field E remains constant.
- Voltage is the work done per unit charge to move it through a distance.
- Here we increase the distance so the work done per unit charge increases.
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