In Fig. 21-24, the particles have charges q1 q2 100 nC and q3 q4 200 nC, and distance a 5.0 cm. What are the (a) x and (b) y components of the net electrostatic force on particle 3?

The force of q1 on q3 is a repulsive force and those of q2 and q4 on q3 are attractive forces. From second diagram at the bottom of the page in the attachment below it can be seen that the force of q1 on q3 is directed vertically downward and has on y component while the force of q4 on q3 is directed to the right and has on x component. The force of q2 on q3 has both x and y components. The full solution can be found in the attachment below.

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mahima6824soni mahima6824soni · Answer 2 · 2022-01-17T10:04:37+01:00

The electric force between two charged bodies at rest is called as electrostatic force.

The values are the following:

The x component of the net electrostatic force = 0.17 N

The y component of the net electrostatic force = 0.045 N

Solution :

For component X

[tex]F_3_y = \dfrac{lql}{4\pi\varepsilon _0q^2 } (\dfrac{-lq_1l+ lq_1l}{\sqrt[2]{2} } )[/tex]

[tex]= (8.99 \times 10^{9} N.m2/ C^{2} \dfrac{(2.0\times10^{-7} )^2} {(0.050 m)^2}(-2 +\frac{2}{2\sqrt{2} } )\\ = 0.17 N[/tex]

X= 0.17 N

For component Y

[tex]F_3_y = \dfrac{lql}{4\pi\varepsilon _0q^2 } (\dfrac{-lq_1l+ lq_1l}{\sqrt[2]{2} } )[/tex]

[tex]= (8.99 \times 10^{9} N.m^2/ C^{2} \dfrac{(1.0\times10^{-7} )^2} {(0.050 m)^2}(-1 +\frac{1}{2\sqrt{2} } )\\ = 0.045 N[/tex]

Y = 0.045 N.

Hence, X= 0.17 N and Y = 0.045 N.

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