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katen-ka-za [31]
3 years ago
9

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

charge density 3.00 uC/m.. . What is the kinetic energy of the sphere when it is 4.50cm from the line of charge if the only force on it is the force exerted by the line of charge?
Physics
1 answer:
JulijaS [17]3 years ago
5 0
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
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Answer: 56.44°

Explanation:

<u>Given:</u>

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<em>Converting to SI Units (m/s):</em>

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<u>Find:</u>

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We have, sinθ = speed of sound / speed of object

               sinθ = v / u

                   θ = sin^-1 (v / u)  

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                   θ = 56.44°

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The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

  • t=0.476v
  • t=1.967v
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<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
  • 20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

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Therefore

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Therefore, the values of voltages at the various times are

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Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

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