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A proton is placed in an electric field of intensity 700 n/

c. what is the magnitude and direction of the acceleration of this proton due to this field?

1 Answer

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F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line

User Douglas Parker
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