Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K|QQ′|d2, where K=14πϵ0, and ϵ0=8.854×10−12C2/(N⋅m2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q1 = -14.0 nC , is located at x1 = -1.655 m ; the second charge, q2 = 31.0 nC , is at the origin (x = 0).

What is (Fnet3)x(Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3q3 = 45.0 nCnC placed between q1q1 and q2q2 at x3x3 = -1.105 mm ?

Your answer may be positive or negative, depending on the direction of the force.

Answer 1

a. 91, x = -165570 (192) +1-105 is towards ai þz 1.655m 92 OD 2 due to a , aa 10 3025 1.22 of force by age. 9,--14'0nc 31.onc aga 45.one x= -1.655m X₂= 715 5 q, is Ve and ag is tve. So attraction, F 9, F3, is the force gi is the end is also the Chiko charges Repel each other) furs q2 Quay Force on az due to a (F34) is Both forces acting aloog -n direction. Fiber & C- Fox = Kqz K939 fa 9 da di? For = 9x109 x 45x10²x [14x10 7 31x101) d. - lx,l-AX31 = 1.655-1-105=0557 [[0.55) + (1.105 d₂ = 1X₂1 = 1.105 m 9x45x109/14 Here 33 con charge 9x45x16" [46.26 + 25.41 sign. Bo, No need to put = 0.29+104 N charge sign during calculating ce + F 1.655- 1.105 0-559 31 ka, d kaga (-a 32 + we are finding drection + the force