A proton and an electron form a dipole separated by 1.5e-10 m along the y-axis centered at the origin. What is the magnitude of
the net electric field at 6.9e-10 m on the y-axis?
1 answer:
Answer:
E = - 7.2 * 10^8 N/C
Explanation:
Attached below is a rough sketch I made of the problem.
The net electric field at the point of consideration is the sum of the electric field due to the electron and the electric field due to the proton.
That is:
E = E(e) + E(p)
Note that Electric field is given by:
E = kq/r^2
Where
k = Coulombs constant
q = charge
r = distance between charge and point of consideration.
ELECTRIC FIELD DUE TO PROTON
The point of consideration is 7.65 * 10^-10 m away from the proton. i.e.
(6.9 * 10^-10) + (0.75 * 10^-10) m
Hence the electric field due to the proton is:
E(p) = (k * e) / (7.65 * 10^-10)^2
Where e = electronic charge.
E(p) = (9 * 10^9 * 1.6022 * 10^-19) / (7.65 * 10^-10)^2
E(p) = 2.46 * 10^9 N/C
ELECTRIC FIELD DUE TO ELECTRON
The point of consideration is 6.15 * 10^-10 m away from the electron, i.e.
(6.9 * 10^-10) - (0.75 * 10^-10)
Hence the electric field due to the electron is:
E(e) = (-k * e) / (6.15 * 10^-10)^2
Where e = electronic charge.
E(e) = -(9 * 10^9 * 1.6022 * 10^-19) / (6.15 * 10^-10)^2
E(e) = -3.18 * 10^9 N/C
NET ELECTRIC FIELD
The net electric field due to the electron and proton will the be:
E = E(e) + E(p)
E = -(3.18 * 10^9) + (2.46 * 10^9)
E = - 0.72 * 10^9 N/C
E = - 7.2 * 10^8 N/C
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