Question

A proton and an electron are moving due east in a constant electric field that also points due east. The electric field has a magnitude of 8.0 × 104 N/C. Determine the magnitude of the acceleration of the proton and the electron.

Answer 1

Electrostatic force is given by,

Fe = q*E

q = charge

E = electric field

i.)

Force on proton,

Fp = q*E

q = +e = 1.60*10^-19 C

E = 8.0*10^4 N/C

By newton's law,

Fnet = m*a

here, m = mass of proton(mp) = 1.67*10^-27 kg

a = acceleration of proton = ap = ??

So, Fp = mp*ap

ap = Fp/mp = (1.60*10^-19)*(8.0*10^4)/(1.67*10^-27)

ap = 7.66*10^12 m/s^2

ii.)

Force on electron,

Fe = q*E

q = -e = -1.60*10^-19 C

E = 8.0*10^4 N/C

By newton's law,

Fnet = m*a

here, m = mass of electron(me) = 9.1*10^-31 kg

a = acceleration of electron = ae = ??

So, Fe = me*ae

ae = Fe/me = (-1.60*10^-19)*(8.0*10^4)/(9.1*10^-31)

ae = -1.41*10^16 m/s^2

magnitude of acceleration = |ae| = 1.41*10^16 m/s^2

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