16.3 COULOMB\'S LAW 589 Example 16.3 continued 0.5 m Now for the force due to ch

Question

16.3 COULOMB'S LAW 589 Example 16.3 continued 0.5 m Now for the force due to charge 3. 8.99 x 10,Nur 2 x (0.20x10"O x (060 x 10°C) = 4.32x 10" Na 432 miN Adding the two force vectors gives the total force F 05 m (0.50 m Figure 16.15 Lacation of point charges in Example 16.3 Strategy The force on q due to q and the force on q2 due are determined separately. After sketching a free body agram, we add the two forces as vectors. Let the distance between charges 1 and 2 be r12 and the distance between charges 2 and 3 be rzs. Findin g x- and y-components: Fm"-F21 sin 3.83mN x t -353nN = som =-1.47 nN Solution Charges 1 and 3 are both positive, but charge 2 is F, is in the -ydirection, so Fax0 and as432mN negative. The forces acting on charge 2 due to charges land Adding components, Fa.-353 nN and F, (147m2 3 are both attractive. Figure 16.16a shows an FBD for +(-4.32 mN)--5.79 mN. The magnitude of P, is charge 2 with force vectors pointing toward each of the other Now find the magnitude of force F21 on q2 duc to q, from Coulomb's law and then repeat the same process to find the From Fig, 16.16c. F, is clockwise from the -y-axis by an magnitude of force Fs2 on q2 due to qy angle The distance between charges 1 and 2 is, from the ngie Pythagorean theorem, Discussion The net force has a direction compatible with the graphical addition in Fig. 16.16b-it has components in the -x and-y-directions From Coulomb's law, Practice Problem 16.3 Find the magnitude and direction of the electric force on Electric Force on Charge 3 .899 x 10°Mm: × (1.2×10. C) x (0.60 x 10"O 1.30 m) charge 3 due to charges I and 2 in Fig. 16.15 -383 x 10" N = 3.83 m1N r Figure 16.16 (a) Free-body diagram showing the directions of forces F, and b) Vectors P and F and dheir s F(c) Finding the direction of F, from its x- and y-components lc) acting on the microscopic building blocks of matter (e.g. s, ions, and electrons), we find that the electric forces are much stron- gravitational forces between them. Only when we put a large aumber of Tecules together to make a massive object can the gravitational force dom- ee mutation occurs only because there is an almost perfect balance between e consider the forces egative charges in a large object, leading to nearly zero net charge.

Explanation / Answer

Solution) From fig

By pythogaras theorem

r12=((r13)^2+(r23)^2)^(1÷2)

r12=((1.2)^2+(0.5)^2)^(1÷2)

r12=1.3m

F12=kq1q2/r12^2

F12=8.99×10^(9)×1.2×10^(-6)×0.60×10^(-6)/(1.3)^2

F12=3.83mN

Since charge q2 is negative charge so it gets attracted towards positive charge on q1 and q3

As seen from figure it acts towards q1 and q3 so it has two components

F23=kq2q3/r23^2

F23=8.99×10^9×0.20×10^(-6)×0.60×10(-6)/(0.5)^2

F23=4.32mN

F2=(F2x^2+F2y^2)^(1÷2)

F21x=-F21sin(theeta)=-3.83×1.20/1.3=-3.53mN

- 've sign indicates negative x direction

F21y=-F21cis(theeta)=-3.83×0.5/1.3=-1.47mN

On substituting we get F2=6.8mN

Angle phi=tan inverse(3.53/5.79)=31

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