Question

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line and is moving directly toward the line with speed 3000 m/s . How close does the proton get to the line of charge? Express your answer in meters.

Answer 1

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm