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lesya [120]
3 years ago
14

A parallel-plate capacitor has circular plates of 7.10 cm radius and 1.26 mm separation. What charge (in nC) will appear on the

plates if a potential difference of 126 V is applied?
Physics
1 answer:
erik [133]3 years ago
6 0

Answer:

13.98 nC

Explanation:

Capacitance depends upon the area of the plates and their distance of separation.

Radius = r = 0.071 m

separation = d = 0.00126 m

C = \frac{\kappa  \epsilon _{o} A }{d}

here κ = 1 and ε₀ = 8.85 ₓ 10⁻¹² SI units , for free space.

Area = A = π r² = 0.0158 m²

C = [( 8.85 ₓ 10⁻¹² ) ( 0.0158) ]÷ (0.00126) = 1.11 x 10⁻¹⁰ F

Charge = Q = C V = ( 1.11 x 10⁻¹⁰ F )(126) = 13.98 nC

                   = 14 nC ( rounded to two significant digits)

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Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

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I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

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EleoNora [17]

Answer:

Revolutions made before attaining angular velocity of 30 rad/s:

θ = 3.92 revolutions

Explanation:

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<h3>Find Torque:</h3>

Torque is the rate of change of angular momentum:

T = \frac{L(final)-L(initial)}{t}\\T = \frac{10.7-0}{8}\\T=1.34 Nm

<h3>Find Angular Acceleration:</h3>

We know that

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α = 1.34/2.2

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<h3>Find Time 't'</h3>

We know that angular equation of motion is:

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θ = 24.6/ 2π

θ = 3.92 revolutions

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