Answer
Elecric field due to uniformly charged with a constant charge density λ. Is given
dE=2kλrdrdE=\frac{2k\lambda}{r}drdE=r2kλdr
Work done is given by
dW=QdEdW=QdEdW=QdE
So total work done
W=∫y1y2Q2kλrdrW=\int^{y_2}_{y_1} Q\frac{2k\lambda }{r}drW=∫y1y2Qr2kλdr
W=2kQλln(y2y1)W=2kQ\lambda \ln{ (\frac {y_2}{y_1})}W=2kQλln(y1y2)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments