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Q .81 - Excercises And Problems

Expert-verified
Found in: Page 746

### Physics For Scientists & Engineers

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

$\frac{2n}{4x+1}\left(1-\frac{d}{l}\mathrm{ln}\left(1+\frac{1}{d}\right)}.$
See the step by step solution

## Step 1: Given information

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:

$V\frac{1}{4\pi {r}_{0}}\frac{q}{r}$

## part(a) step 1: given information

(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,${\lambda }_{0}=\frac{2Q}{L}$

## part(b) step 1: given information

(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

$V=\int dV\phantom{\rule{0ex}{0ex}}=\frac{2}{4\pi xL}{\int }_{0}^{L}\frac{x}{d+x}dx\phantom{\rule{0ex}{0ex}}=\frac{2O}{4\pi +{L}^{2}}{\int }_{0}^{L}\left(1-\frac{d}{d+x}\right)dx\phantom{\rule{0ex}{0ex}}=\frac{2Q}{4\pi ·}\left\{1-\frac{d}{l}\mathrm{ln}\left(1+\frac{L}{d}\right)\right\}$