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N76 [4]
3 years ago
9

The potential difference between two parallel plates is 227 V. If the plates are 6.8 mm apart, what is the electric field betwee

n them? O S.0x10 N/C O 28 x 10* N/C O 4.1 x 10 N/C O 3.3 x 10 N/C
Physics
1 answer:
nalin [4]3 years ago
3 0

Answer:

E=3.3\times 10^4N/C

Option D is the correct answer.

Explanation:

Electric field, E is the ratio of potential difference and distance between them.

Potential difference, V = 227 V

Distance between plates = 6.8 mm = 0.0068 m

Substituting,

         E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C

Option D is the correct answer.

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A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be
vodomira [7]

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

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ii )

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in the second case potential difference = Q / 2C

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iii ) electric field = potential diff / plate separation

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

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So electric field remains unchanged .

iv)

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In the second case energy stored = Q² / 2x2C

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4 0
4 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
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V1 = 1.5 l

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6 0
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A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to
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Answer:

v = 24 m/s, rightwards

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Given that,

The mass of TBT explosive = 5 kg

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