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andrew11 [14]
3 years ago
13

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

4.35 cm. If a third charge of 5.00nC, is placed at the point x=3.10cm, y=3.80cm find the x and y components of the total force exerted on the charge by two other charges
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis is the net horizontal force

F_v=80062.47\times 10^9 attractive toward +y axis is the net vertical force

Explanation:

Given:

  • charge at origin, Q_0=-3.35\times 10^{-6}\ C
  • magnitude of second charge, Q_2=2.05\times 10^{-6}\ C
  • magnitude of third charge, Q_3=5\times 10^{-6}\ C
  • position of second charge, (x_2,y_2)\equiv(0,4.35)\ cm
  • position of third charge, (x_3,y_3)\equiv(3.1,3.8)\ cm

<u>Now the distance between the charge at at origin and the second charge:</u>

d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}

d_2=\sqrt{(0-0)^2+(4.35-0)^2}

d_2=0.0435\ m

<u>Now the distance between the charge at at origin and the third charge:</u>

d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}

d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}

d_3=0.04904\ m

<u>Now the force due to second charge:</u>

F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}

F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}

F_2=32175.98\times 10^9\ N attractive towards +y

<u>Now the force due to third charge:</u>

F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}

F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}

F_3=61748.38\times 10^9\ N attractive

<u>Now the its horizontal component:</u>

F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9

F_{3h}=39065.298\times 10^9\ N attractive toward +x axis

<u>Now the its vertical component:</u>

F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9

F_{3v}=47886.49\times 10^9\ N upwards attractive

Now the net vertical force:

F_v=F_{3v}+F_2

F_v=47886.49\times 10^9+32175.98\times 10^9

F_v=80062.47\times 10^9

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