- #1

stunner5000pt

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Hvae a look at the diagram

Caluclate the force on the point charge q, due to a uniformly charged rod of length L a distance x from the point charge q. Discuss the limit when L approaches infinity with lambda = Q/L fixed.

[tex] Q = \lambda L [/tex]

[tex] dQ = \lambda dL [/tex]

Force of dQ on the point charge q is given by

[tex] dF = \frac{1}{4 \pi \epsilon_{0}} \frac{qdQ}{(L+x-s)^2} [/tex]

no Y components since the rod is thin. SO this force is the total force in the horizontal direction only.

[tex] F = \int dF = \int_{s=0}^{s=L} \frac{1}{4 \pi \epsilon_{0}} \frac{q \lambda ds}{(L+x-s)^2} [/tex]

[tex] F = \frac{q \lambda L}{4 \pi \epsilon_{0}} \frac{1}{x(L+x)}[/tex]

[tex] F = \frac{Qq}{4 \pi \epsilon_{0} x(L+x)} [/tex]

now for the limit where L -> infinity

i used L'Hopital's Rule and got the answer to be zero. But i find it hard to believe that that is the case. I Would think that this has something to di wth the limtis of integration being s=0 to s=infinity

im not sure however... do help

thank you in advance for your help!

Caluclate the force on the point charge q, due to a uniformly charged rod of length L a distance x from the point charge q. Discuss the limit when L approaches infinity with lambda = Q/L fixed.

[tex] Q = \lambda L [/tex]

[tex] dQ = \lambda dL [/tex]

Force of dQ on the point charge q is given by

[tex] dF = \frac{1}{4 \pi \epsilon_{0}} \frac{qdQ}{(L+x-s)^2} [/tex]

no Y components since the rod is thin. SO this force is the total force in the horizontal direction only.

[tex] F = \int dF = \int_{s=0}^{s=L} \frac{1}{4 \pi \epsilon_{0}} \frac{q \lambda ds}{(L+x-s)^2} [/tex]

[tex] F = \frac{q \lambda L}{4 \pi \epsilon_{0}} \frac{1}{x(L+x)}[/tex]

[tex] F = \frac{Qq}{4 \pi \epsilon_{0} x(L+x)} [/tex]

now for the limit where L -> infinity

i used L'Hopital's Rule and got the answer to be zero. But i find it hard to believe that that is the case. I Would think that this has something to di wth the limtis of integration being s=0 to s=infinity

im not sure however... do help

thank you in advance for your help!

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