1. (a) Three point charges, A = 1.55 µC, B = 6.65 µC, and C = 3.95 µC, are locat
ID: 1529767 • Letter: 1
Question
1.
(a) Three point charges, A = 1.55 µC, B = 6.65 µC, and C =
3.95
µC, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 1.55 µC charge.
(b) How would the electric field at that point be affected if the charge there were doubled?
The magnitude of the field would be halved.The field would be unchanged. The magnitude of the field would double.The magnitude of the field would quadruple.
Would the magnitude of the electric force be affected?YesNo 2.Three charges are at the corners of an equilateral triangle, as shown in the figure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let
q1 = 7.50 C,
q2 = 7.00 C,
and
q3 = 4.50 C.)
magnitude N/C direction ° below the +x-axis BAC 0.500 m 60.0 AAC CACExplanation / Answer
1)
a) x-component of resultant electrcic field,
Ex = -|EB|*cos(60) + EC
= -k*qB/L^2)*cos(60) + k*qC/L^2
= -(9*10^9*6.65*10^-6/0.5^2)*cos(60) + 9*10^9*3.95*10^-6/0.5^2
= 2.25*10^4 N/c
y-component of resultant electrcic field
Ey = -|EB|*sin(60)
= -(9*10^9*6.65*10^-6/0.5^2)*sin(60)
= -20.7*10^4 N/c
E = sqrt(Ex^2 + Ey^2)
= sqrt(2.25^2 + 20.7^2)*10^4
= 20.8*10^4 N/c <<<<<<<<-------------------Answer
direction : theta = tan^-1(Ey/Ex)
= tan^-1(20.7/2.25)
= 83.8 degrees below +x axis <<<<<<<<-------------------Answer
b) The field would be unchanged
c) yes.
2) at mid point,
Ex = k*(q2 + q3)/(L/2)^2
= 9*10^9*(7 + 4.5)*10^-6/0.25^2
= 1.66*10^6 N/c
Ey = -k*q1/sqrt(L^2 - (L/2)^2)
= -9*10^9*7.5*10^-6/sqrt(0.5^2 - 0.25^2)
= -1.56*10^5 N/c
E = sqrt(Ex^2 + Ey^2)
= sqrt(1.66^2 + 0.156^2)*10^6
= 1.67*10^6 N/c <<<<<<<<-------------------Answer
direction : theta = tan^-1(Ey/Ex)
= tan^-1(0.156/1.66)
= 5.37 degrees below +x axis <<<<<<<<-------------------Answer
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