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2 years ago

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Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

e plate #2 has a charge −2 Q. Find the magnitude of the electric field at a point P in the gap. Assume that the separation between the plates is very small compared to the dimensions of the plates.

1 answer:

Karo-lina-s [1.5K]2 years ago

5 0

Answer:

$E = \frac{3Q}{2A\epsilon_0}$

Explanation:

By Gauss Law for electric field:

$E = \frac{\sigma}{2\epsilon_0}$

Where $\sigma$ is the charge density Q/A. Since we have 2 parallel plates with different charge, the electric field at point P in the gap would be the sum of 2 field

$E = E_1 + E_2$

$E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}$

$E = \frac{3Q}{2A\epsilon_0}$

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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

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Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

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Principle of Angular Impulse and Momentum

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$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

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