Point charge q 1 = +2.51 ? C is at point A (-22.65 cm,0), and point charge q 2 ,
ID: 1316252 • Letter: P
Question
Point charge q1 = +2.51 ?C is at point A (-22.65 cm,0), and point charge q2, carrying an equal and opposite charge, is at point B (+22.65 cm,0).
a) Find the magnitude of the electric field at the origin.
b) Give the direction of the electric field at the origin.
into the page
out of the page
upward
downward
left, toward the negative charge
right, toward the positive charge
zero
c) An electron (q = 1.60x10-19, m = 9.11x10-31 kg) is placed at the origin and released. Find the magnitude of the acceleration the electron would experience, at the instant it is released, due to the two charges.
a = m/s2
d) Give the direction of the force on the electron.
left, toward the negative charge
right, toward the positive charge
zero
upward
downward
into the page
out of the page
Explanation / Answer
E1 = k*q1/r^2 towards -ve charge
E2 = k*q2/r^2 towards -ve charge
E = E1 +E2
E = k*q1/r^2 + k*q2/r^2
q1 = q2 = 2.51e-6 C
E = (2*9e9*2.51e-6)/(0.2265*0.2265) = 8.806*10^5 N/C
from positive to negative
left, toward the negative charge
c)
F = E*q = 8.806e5*1.6e-19 = 1.40896e-13 N
a = F/m = 1.40896e-13/9.11e-31 = 1.55*10^17 m/s^2
d) from negative to positive
right, toward the positive charge
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