Answer:
Acceleration of proton will be [tex]-9.197\times 10^{12}m/sec^2[/tex]
Explanation:
We have given surface charge density of an infinite charged plate [tex]\sigma = -1.70\times 10^{-6}C/m^2[/tex]
Electric field due to infinite sheet charge is given by [tex]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[/tex]
Charge in proton is given by [tex]e=1.6\times 10^{-19}C[/tex]
So force on proton [tex]F=qE=1.6\times 10^{-19}\times -9.6\times 10^4=-15.36\times 10^{-15}N[/tex]
Mass of proton [tex]m=1.67\times 10^{-27}kg[/tex]
According to newtons second law force F = mass × acceleration
So [tex]-15.36\times10^{-15}=1.67\times 10^{-27}a[/tex]
[tex]a=-9.197\times 10^{12}m/sec^2[/tex]
