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An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s2. What are the magnitude and direction of the electric field.

• Physics -

Assuming I did it correctly:

F=ma

mass = the mass of the electron = 9.11 * 10^-31 kg

a= 115 m/s

F will equal 1.05 * 10^-28 Newtons (N)

Next, F=qE

q equals the charge of the electron, or "point charge".
q=1.6 * 10^-19 C

Solve for E, so E=F/q

plug in your values, and E will equal 6.56 *10^-10 N/C

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