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:
Different and better?
Explanation:
i dont think that helps lol
Answer:
Explanation:
Given that
F=2x³
Work is given as
The range of x is from x=0 to x=D
W=-∫f(x)dx
Then,
W=-∫2x³dx from x=0 to x=D
W=- 2x⁴/4 from x=0 to x=D
W=-2(D⁴/4-0/4)
W=-D⁴/2
W=1/2D⁴
The correct answer is F
Answer:
B it decreases
Explanation:
the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy
