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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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(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from r
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Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = 13\times 0.0254=0.3302\ m

r = Radius = \frac{d}{2}=\frac{0.3302}{2}=0.1651\ m

\omega_f = Final angular velocity = 80\times\frac{2\pi}{60}=8.37758\ rad/s

\omega_i = Initial angular velocity = 0

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Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2

Tangential acceleration is given by

a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2

Direction is given by

\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°

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