3. –/4 points Notes Question: SerCP8 15.P.024. .Question part Points Submissions

Question

3. –/4 points Notes Question: SerCP8 15.P.024. .Question part Points Submissions 1 2 3 4
0/1 0/1 0/1 0/1
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(a) Three point charges, A = 2.2 µC, B = 7.5 µC, and C = -3.9 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 2.2 µC charge.
magnitude 1 N/C
direction 2° (counterclockwise from the +x-axis)

(b) How would the electric field at that point be affected if the charge there were doubled?

3
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.
.

(c) Would the magnitude of the electric force be affected?

4
no
yes
.

Explanation / Answer

(a) To find the force on A by B and C, just use coulomb's law: F1 is the force on A by B F2 is the force on A by C F1=k(2.2*10^-6)*(7.5*10^-6)/r^2 F2=k(2.2*10^-6)*(-3.9*10^-6)/r^2 (we use *10^-6 because we are given µC) Then you must find the x an y components of your forces, add them vectorially, and find the magnitude of the resultant vector: y component of force 1 = F1sin(60), x component = F1cos(60) y component of force 2 =F2sin(-60), x component = F2cos(-60) (60 degrees because its an equilateral triangle) Resultant force = magnitude = sqrt(x^2 + y^2)

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