Respuesta :
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
[tex]V=-2K\lambda ln\left (\frac{r'}{r} \right )[/tex]
where, K = 9 x 10^9 Nm^2/C^2
W = q V
By use of work energy theorem
Work = change in kinetic energy
[tex]qV = 0.5m(u^{2}-v^{2})[/tex]
By substituting the values, we get
[tex]V=\frac{mu^{2}}{2q}[/tex]
[tex]-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}[/tex]
[tex]- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }[/tex]
[tex]- 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} }[/tex]
[tex]- ln\left ( \frac{r'}{r} \right )=0.35[/tex]
[tex]\frac{r'}{r} =e^{-0.35}[/tex]
[tex]\frac{r'}{r} =0.7047[/tex]
r' = 14.5 x 0.7047 = 10.22 cm
