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timama [110]
3 years ago
10

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 a

t 1.90 x 10^6 m/s . How far does the proton travel before reaching its turning point?
Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0

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

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