(Figure 1) shows five electric charges. Four charges with themagnitude of the charge 2.0 nC...

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(Figure 1) shows five electric charges. Four charges with themagnitude of the charge 2.0 nC...

(Figure 1) shows five electric charges. Four charges with the magnitude of the charge 2.0 nC form a square with the sizea = 3.0 cm. Positive charge with the magnitude of q = 7.0 nC is placed in the center of the square.

-2.0 nC 2.0 nC 2.0 nC 2.0 nC

What is the direction of the force on the 7.0 nC charge in the middle of the figure due to the four other charges?

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Answer 1

Answer #1

due to left-upper (-2nC) :

F1 = (kq1q2/d^2) (cos@i + sin@j)

F1= (9 x 10^9 x 2 x 10^-9 x 7 x 10^-9/ (0.015^2 + 0.015^2)) ( - cos45i + sin45j)

F1 = (-2.8e-4i + 2.8e+4 ) N

due to left-lower charge:

F2 = (9 x 10^9 x 2 x 10^-9 x 7 x 10^-9/ (0.015^2 + 0.015^2)) ( - cos45i - sin45j)

F2 = (-2.8e-4i - 2.8e+4 ) N

due to right-lower charge:

F3 = (9 x 10^9 x 2 x 10^-9 x 7 x 10^-9/ (0.015^2 + 0.015^2)) ( - cos45i + sin45j)

F3 = (-2.8e-4i + 2.8e+4 ) N

due to right-upper charge:

F2 = (9 x 10^9 x 2 x 10^-9 x 7 x 10^-9/ (0.015^2 + 0.015^2)) ( - cos45i - sin45j)

F2 = (-2.8e-4i - 2.8e+4 ) N

Fnet = F1 + F2 + F3 + F3

= -4 x 2.8 x 10^-4 (i )

= 1.12 x 10^-3 (-i) N

Magnitude = 1.12 x 10^-3 N Or 1.12 mN

dierction - to the left

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Answer 2

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