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>
Answer:
The rate at which energy is transferred is called power and the amount of energy that is usefully transferred is called efficiency.
To solve this problem, we are going to use the formula for
work which is Fd where x and y are measured separately.
X direction: W = 13.5 x 230 = 3105 Joules
Y direction: W = -14.3 x -165 = 2360 Joules
So the total work is getting the sum of the two: 3105 + 2360
= 5465 Joules
Longitude was. Determining longitude requires knowing the exact time of day, which was difficult prior to modern clocks. The source book below tells the story of Englishman John Harrison's life-long pursuit of building a reliable clock and its importance to navigation.
