Question

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the rod's left end.



What is λ0 in terms of Q and L?



Express your answer in terms of some or all of the variables Q, L.



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



Express your answer in terms of Q, L, d, π, ϵ0.

Answer 1

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

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

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

Answer 2

The constant λ₀ in the linear charge density in terms of Q and L is λ₀=2Q/L

And the potential difference at the given point is  $\frac{Q}{4\pi \epsilon_0 L^2}[L +d\ln(\frac{d}{d+L})]$

Linear charge density and potential:

Given that the linear charge density is :

λ(x) = λ₀x/L

The total charge is given by:

$Q=\int\limits^L_0 {(\lambda_0x/l)} \, dx\\\\Q=\frac{\lambda_0}{2L}[x^2]_0^L\\\\Q=\frac{\lambda_0L}{2}$

λ₀=2Q/L

The electric potential on the axis at distance d left of the rod's left end:

$dV=\frac{kdQ}{d+x}\\\\V=\frac{2KQ}{L^2}\int\limits^L_0 {\frac{x}{d+x} } \, dx \\\\V=\frac{2KQ}{L^2}\int\limits^L_0 {(1-\frac{d}{d+x} }) \, dx\\\\V=\frac{2KQ}{L^2}[x-d\ln(d+x)]_0^L\\\\V=\frac{Q}{4\pi \epsilon_0 L^2}[L +d\ln(\frac{d}{d+L})]$

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