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Anettt [7]
3 years ago
10

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching e

ach other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease or stay the same?i) Cii) Qiii) E between the platesiv) delta V
Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

C=\frac{\epsilon_0.A}{d}

\therefore C\propto \frac{1}{d}

<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>

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:

E=\frac{\sigma}{\epsilon_0}

where:

charge density, \sigma=\frac{Q}{A}

<em>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.</em>

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.
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<h3><u>Full question:</u></h3>

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