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4 years ago

Two parallel conducting plates are separated by 4.00 cm. The electric field strength between the two plates is 5.70×10⁴ V/m.

Physics

1 answer:

Nookie1986 [14]4 years ago

6 0

Answer:

2280 V

Explanation:

distance between the plates, d = 4 cm = 0.04 m

Electric field strength, E = 5.7 x 10^4 V/m

The formula for potential difference is given by

V = E x d

V = 5.7 x 10^4 x 0.04

V = 2280 V

thus, the potential difference between the plates is 2280 V.

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The total current in the wire is obtained by integrating

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$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

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3 years ago

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