Question

a. Use the methods of Chapter 25 to find the potential at distance $x$on the axis of the charged rod shown in FIGURE P26.43.

b. Use the result of part a to find the electric field at distance $x$on the axis of a rod

Short answer

a. The potential at distance$x$on the axis of the charged rod$\frac{kQ}{L}In\left|\frac{x+L/2}{x-L/2}\right|$.

b. The electric field at distance$x$on the axis of a rod $E\left(x\right)=\frac{kQ}{x^2-L^2/4}$.

Step-by-step solution

Part (a) step 1: Given information

We need to find the potential at distance $x$on the axis of the charged rod shown in FIGURE $P26.43$.

Part (a) step 2: Simplify

The potential in the integral form is given as:

$V\left(x\right)=k{\int }_{L/2}^{L/2}\frac{Q/Ldr}{r}=\frac{kQ}{L}{\int }_{-L/2}^{L/2}\frac{di}{x+i}$,

where $i$is a dummy variable for the integration and $x$denotes the position at which we calculate the potential. The solution to this integral is

$V\left(x\right)=\frac{kQ}{L}{\left(In\left|x+i\right|\right)}_{-L/2}^{L/2}$,

Hence, it is

$V\left(x\right)=\frac{kQ}{L}In\left|\frac{x+L/2}{x-L/2}\right|$

Part (b) step 1: Given information

We need to find the the electric field at distance $x$on the axis of a rod.

Part (b) step 2: Simplify

Calculating the electric field strength using part a.

$E\left(x\right)=-\frac{dV}{dx}$

Now, we need to derive the expression we just found. The derivative, according to the constants, will be

$\frac{d}{dx}In\frac{x+L/2}{x-L/2}=\frac{1}{{\displaystyle \frac{x+L/2}{x-L/2}}}·\frac{d}{dx}\frac{x+L/2}{x-L/2}$

After careful derivation, one would obtain

$-\frac{L}{x^2-L^2/4}$

Considering this, the electric field strength will be given by

$E\left(x\right)=-\frac{KQ}{L}-\frac{L}{x^2-L^2/4}$,

which we can be simplified to

$E\left(x\right)=\frac{kQ}{x^2-L^2/4}$