Question

The surface charge density on an infinite charged plane is −1.70 x 10^−6 C/m^2 . A proton is shot straight away from the plane at 1.90 x 10^6 m/s . How far does the proton travel before reaching its turning point?

Answer 1

Answer:

Acceleration of proton will be $-9.197\times 10^{12}m/sec^2$

Explanation:

We have given surface charge density of an infinite charged plate $\sigma = -1.70\times 10^{-6}C/m^2$

Electric field due to infinite sheet charge is given by $E=\frac{\sigma }{2\epsilon _0}=\frac{-1.7\times 10^{-6}}{2\times 8.85\times10^{-12}}=-0.096\times 10^6=-9.6\times 10^4N/C$

Charge in proton is given by $e=1.6\times 10^{-19}C$

So force on proton $F=qE=1.6\times 10^{-19}\times -9.6\times 10^4=-15.36\times 10^{-15}N$

Mass of proton $m=1.67\times 10^{-27}kg$

According to newtons second law force F = mass × acceleration

So $-15.36\times10^{-15}=1.67\times 10^{-27}a$

$a=-9.197\times 10^{12}m/sec^2$