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Free_Kalibri [48]
2 years ago
9

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

lates is 365 V , and the plate separation is 0.200 mm .
Physics
1 answer:
PSYCHO15rus [73]2 years ago
5 0

Answer:

Energy density will be 14.73 J/m^3

Explanation:

We have given capacitance C=225\mu F=225\times 10^{-6}F

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm 0.2\times 10^{-3}m

We know that there is relation between electric field and potential

E=\frac{V}{d}, here E is electric field, V is potential and d is separation between the plates

So E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C

Energy density is given by E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3

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Answer:

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3 0
2 years ago
A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

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length of the string r = 70 cm = 0.7 m

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\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

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6 0
2 years ago
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sveticcg [70]

Answer:

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Answer:

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6 0
3 years ago
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borishaifa [10]

Hi there!

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7 0
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