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Alex777 [14]
3 years ago
7

The electric field in the region between the plates of a parallel plate capacitor has a magnitude of 6.3 105 V/m. If the plate s

eparation is 0.56 mm, determine the potential difference between the plates.
Physics
1 answer:
xz_007 [3.2K]3 years ago
4 0

Answer:

3.528×10² V.

Explanation:

potential difference: This is the work done when one coulomb of charge moves from one point to another in an electric field. The S.I unit of potential difference is volt. The formula of potential difference is given as,

V = kq/r..................... Equation 1

And

E = kq/r² .................. Equation 2

Comparing equation 1 and equation 2,

V = E×r............................. Equation 3

Where V = potential difference, E = Electric field between the plate of the capacitor, r = distance between the plate.

Given: E = 6.3×10⁵ V/m, r = 0.56 mm = 0.00056 m.

Substitute into equation 3,

V = 6.3×10⁵×0.00056

V = 3.528×10² V.

Hence the potential difference of the plate = 3.528×10² V.

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Answer:

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Explanation:

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Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

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