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 = -15.5 nC , is located at x1 = -1.715 m ; the second charge, q2 = 30.0 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 54.0 nC placed between q1 and q2 at x3 = -1.105 m ?

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

Answer 1

Given : 2, = -15-5 nl x = -1.715m 21LLLPPPPPP 22=30nd *2=0.00 23= Sunc X3 =-11osm 22=3ont 9. =-15.5m 23=54nc *13 105 25000 X =-1.715 Fnet = fio + F23 Fiz = K2, 23 9x109x (-15-5x109) (S4x10g (8,332 = · 715 - 1105) = 9x(1155) (54)×109 10.61) = 9+ (-15.5) xSixioa 0.3721 5 - 20244.5 x1ö9 N F3 = -2.024 x 10-SN

Fz3= . = K22 23 (X23 32 9x109x30x10x suxio-e (1.105) = 9x30x 54X10-9 1.221 3 11941-03x10-9N 1.19x10'' N Fnet = Fiza F23 = -2:04x0wt 119x10-5 = -0.834X10 N - Hence Fret = -0.83lxusn towards - ve x Osê