Question

A proton is launched from an infinite plane of charge with surface charge density -1.10×10-6 C/m2. If the proton has an initial speed of 2.40×107 m/s, how far does it travel before reaching its turning point? 48.4 m 96.7 m 2.02×10-6 m 8.74×10-7 m

Answer 1

The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

$E=\frac{\sigma}{2\epsilon_0}$

where in this case,

$\sigma = -1.10\cdot 10^{-6} C/m^2$ is the surface charge density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

Substituting,

$E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C$

And the direction is towards the plane (because the charge is negative).

The electric force on the proton due to this field is

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

Substituting,

$F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N$

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

$a=\frac{F}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is the proton mass

Substituting,

$a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2$

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

$v^2-u^2=2as$

where

v = 0 is the final velocity

$u=2.40\cdot 10^7 m/s$ is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m$

Therefore, the closest answer is 48.4 m.

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