(adsbygoogle = window.adsbygoogle || []).push({}); Particle of mass, m, charge q>0 is projected toward a nucleus of charge +Q assumed to be at fixed postiion/

a) If the aim is perfect then how close to the center of the nucleus will the particle reach when it comes to a halt?

[tex] q(V_{f} - V{i}) = K [/tex] Vi = 0

[tex] q \frac{1}{4 \pi \epsilon_{0}} \frac{qQ}{r} = K [/tex]

[tex] r = \frac{Qq^2}{4 \pi \epsilon_{0} K} [/tex]

am i correct with the formula for V?

With a particular imperfect aim the particle reaches twice the distnace as above. Calculate the initial velocity.

So i imply plug r = 2r into the first answer and break K = 1/2 mv^2 and solve for v?

A charge per unit length lambda is distributed uniformly along a thin rod of length L.

a) Determine the potential at a point P at a distnace y from one end of the rod, in line with the rod.

so [tex] V = \frac{1}{4 \pi \epsilon_{0}} \int_{s=0}^{s=L} \frac{dq}{y+s} [/tex] [tex]

dq = \lambda ds [/tex]

and [tex] V = \frac{1}{4 \pi \epsilon_{0}} (\ln{y+L} - \ln{y}) [/tex]

[tex] V = \frac{Q}{4 \pi \epsilon_{0} L} \ln{\frac{y+L}{y}} [/tex]

b)Use the result to determine the resulting electric field

Since the rod is thin no X or z components

[tex] E = \frac{\partial V}{\partial y} = \frac{Q}{4 \pi \epsilon_{0} y(y+L)} [/tex]

am i correct?

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# Homework Help: Two questions on electric potential

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