a proton is released from rest in a uniform electric field with a strength 500n/c. find acceleration and speed of proton when it has moved half a meter
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electric field strength , E = 500 N/C
charge on proton, e = +1.6 × 10^-19 C
distance covered by proton, s = 0.5 m
we know, F = qE
and also from Newton's 2nd law, F = ma
so, ma = qE
and a = qE/m , where m is mass of proton.
a = (1.6 × 10^-19 × 500)/1.67 × 10^-27
a = 4.79 × 10^10 m/s²
hence, acceleration of proton is 4.79 × 10^10 m/s²
use formula, v² = u² + 2as
or , v² = 2 × 4.79 × 10^10 × 0.5
hence, v = 2.18 × 10^5 m/s
charge on proton, e = +1.6 × 10^-19 C
distance covered by proton, s = 0.5 m
we know, F = qE
and also from Newton's 2nd law, F = ma
so, ma = qE
and a = qE/m , where m is mass of proton.
a = (1.6 × 10^-19 × 500)/1.67 × 10^-27
a = 4.79 × 10^10 m/s²
hence, acceleration of proton is 4.79 × 10^10 m/s²
use formula, v² = u² + 2as
or , v² = 2 × 4.79 × 10^10 × 0.5
hence, v = 2.18 × 10^5 m/s
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