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Q .81 - Excercises And Problems

Expert-verifiedFound in: Page 746

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

A thin rod of length L and total charge Q has the nonuniform CALC linear charge distribution $\lambda \left(x\right)={\lambda}_{0}x/L$, where x is measured from the rod's left end.

a. What is${\lambda}_{0}$ in terms of Q and L ?

b. What is the electric potential on the axis at distance d left of the rod's left end?

The constant i. comes out to be $\frac{2n}{i}$. whereas the potential at the specified point is found to be

$\frac{2n}{4x+1}\left(1-\frac{d}{l}\mathrm{ln}\left(1+\frac{1}{d}\right)\right\}.$Given:

A thin rod of length L and having a total charge of q, has the linear charge distribution$\lambda \left(x\right)=\frac{\lambda x}{L}$, where x is measured from the left-end of the rod.

Formula used:

- The electric potential at a distance r from a point charge q, is givisn as:

$V\frac{1}{4\pi {r}_{0}}\frac{q}{r}$(a) If we consider a small segment of length$\mathrm{dx}$ at a distance x from left end, then charge on this segment,

$dq=\lambda dx\phantom{\rule{0ex}{0ex}}=\frac{{\lambda}_{0}}{L}xdx$Hence, the total charge on the rod,

$Q={\int}^{Q}dq\phantom{\rule{0ex}{0ex}}=\frac{{\lambda}_{0}}{L}{\int}_{0}^{L}xdx\phantom{\rule{0ex}{0ex}}=\frac{{\lambda}_{0}L}{2}$Therefore,${\lambda}_{0}=\frac{2Q}{L}$

(b) Again because of the small element of charge dq, the small potential at P,

Hence, the potential at point P, due to the entire rod is given as

$V=\int dV\phantom{\rule{0ex}{0ex}}=\frac{2}{4\pi xL}{\int}_{0}^{L}\frac{x}{d+x}dx\phantom{\rule{0ex}{0ex}}=\frac{2O}{4\pi +{L}^{2}}{\int}_{0}^{L}\left(1-\frac{d}{d+x}\right)dx\phantom{\rule{0ex}{0ex}}=\frac{2Q}{4\pi \xb7}\left\{1-\frac{d}{l}\mathrm{ln}\left(1+\frac{L}{d}\right)\right\}$94% of StudySmarter users get better grades.

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