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In case of parallel plate capacitor, if the place separation is doubled and plate area is halved, the capacitance becomes A)4 times B)one half C)one forth D)zero

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samuelonum1

Answer:

C)one forth

Explanation:

  The capacitance of a capacitor is defined as the charge stored per unit potential change.

using the expression for the capacitance

[tex]C= \frac{eA}{d}[/tex]

where e=  electrostatic constant

A = the area of the plate

d = separation distance.

Given the problem statement

1.  If the place separation is doubled

2. plate area is halved.

 effecting the conditions in our expression for the capacitance we have

[tex]C= \frac{e\frac{A}{2} }{2d}\\\\2dC=e\frac{A}{2}[/tex]

cross multiplying we have

[tex]4dC=eA[/tex]

dividing both sides by 4d we have

[tex]C=\frac{eA}{4d}[/tex]

The capacitance is 1/4 the initial value

The capacitance should become C)one forth

The capacitance of a capacitor:

It could be defined by the charge stored per unit with respect to the potential change.

Here the following expression should be used.

[tex]C = eA \div d[/tex]

Here e=  electrostatic constant

A = the area of the plate

d = separation distance.

So in the case when place separation is doubled and plate area is halved

So,

[tex]C = \frac{e\frac{A}{2} }{2d}\\\\2Dc = e\frac{A}{2}\\\\ 4Dc = eA[/tex]

Now if we divide both the sides by 4d so

[tex]C = \frac{eA}{4D}[/tex]

Learn more about plate here: https://brainly.com/question/15612606

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