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Makovka662 [10]
3 years ago
11

A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be

tween the plates is halved, what will happen to (i) Charged stored by capacitor (ii) Potential difference across it. (iii) Electric filed strength between the plates. (iv) Energy stored by the capacitor.
Physics
1 answer:
vodomira [7]3 years ago
4 0

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d  x C )

So electric field remains unchanged .

iv)

energy stored in first case = Q² / 2C

In the second case energy stored = Q² / 2x2C

so energy stored becomes half .

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Answer:

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

Explanation:

Given:

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Force = Mass × gravity acceleration.

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So,

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force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

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When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
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The weight of any object, right now, is

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A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If
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The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.

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We also know the capacitance of a parallel-plate capacitor is given by:

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If we substitute C:

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We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.

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Answer:

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