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neonofarm [45]
3 years ago
15

Recall that there is a uniform electric field inside a capacitor. Recall also that in general, LaTeX: V=-\int{\vec{E}\cdot\vec{d

l}}V = − ∫ E → ⋅ d l →. In regions of uniform field, this simplifies to LaTeX: V=|\vec{E}|dV = | E → | d, where d is the distance traveled parallel to the electric field. A capacitor with a gap of 0.5 mm has a potential difference from one plate to another of 22 V. What is the magnitude of the electric field within the capacitor? Give your answer in units of kV/m.
Physics
1 answer:
My name is Ann [436]3 years ago
7 0

Answer:

E = 44 kV / m

Explanation:

In a capacitor the potential difference is

 

         ΔV = -E x

         E = - ΔV / x

Let's reduce to the SI system

         x = 0.5 mm = 0.5 10⁻³ m

Let's calculate

         E = 22 / 0.5 10⁻³

         E = 44 10³ V / m

         E = 44 kV / m

Note that this electric field is uniform

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The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
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Explanation:

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R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

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3 years ago
Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h
vladimir1956 [14]

Answer:

Explanation:

Given

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T=m_b\times a -----2

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F=\left ( m_a+m_b\right )a

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Jobisdone [24]

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3 years ago
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