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Three charges (q, 5.1 u, 92- -3.9 uC, and q, - 2pC) are located at the 1) What is F,. the value of the x-component of the net force on q,? 2) What is Fs., the value of the y-component of the net force on q:? vertices of an equilateral triangle with side d 7.4 cm as shown. 93 91 9a 3) A charge q. -2 uC is now added as shown. What is F2.x, the x-component of the new net force on q? 4) What is Fav, the y-component of the new net force on q,? 5) What is F, the x-component of the new net force on q,?
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2) Same as above we will do the same calculation by just replacing cos with sin 85.1 × 10-6 × 2 × 10-6 0.0742 Fiy 9 x 10 sin(60°)14.5181 similarly we can write the x component of F2 in the following way 8-19 10-6 2x 10-6 0.0742 Fy 9 x 10 sin(60°)--11.1021 The net force will be Fty F2y 3.41602 N 3) Now we are introducing a fourth charge of magnitude 4μ C The calculations for the x components will remains the same 5.1 x (-3.9) x 1032.6899 Fiz = 9 x 109 0.0742 92x (-3.9) 10-12 0.0742 cos(60°)--6.40979 9 2 x (-3.9) x 10-12 0.0742 Fe = 9 10 cos(60°)--6.40979 The net force will be a sum of all these three forces as FN =-32.6899 +ー6.40979 +-6.40979 =ー45.5095 (10)

4) To find the y component of the forces we will just replace the cos by sin and keep in mind that the force due to charge carry a negative sign as it is acting in-ve y direction. wil А,-Эк 109 5.1x(-3.9) 0.0742 10-12 F3,--8x 1092x (-3.9) x 10-12 (12) 11.1021 0.0742一一-sin(60°) 2x (-3.9) × 10 12 h-9 × 109 sin(60°) 11.1021 (13) 0.0742 so here the net force will be zero

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