Question

An electric point charge of Q1 = 7.85 nC is placed at the origin of the real axis. Another point charge of Q2 = 3.67 nC is placed at a position of p = 2.88 m on the real axis. At which position can a third point charge of q = -3.47 nC be placed so that the net electrostatic force on it is zero? Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?

Answer 1

x = 1.71 m

x = - 9.1068 m

Step-by-step explanation:

from figure 1

[/tex]f_1 = f_2[/tex]

`(kqQ_1)/(X^2) =(kqQ_2)/((2.88-X)^2)`

`(Q_1)/(X^2) =(Q_2)/((2.88-X)^2)`

FROM DATA GIVEN

`(7.85)/(x^2) = (3.67)/((2.88-x)^2)`

solving for x we get

2.88 -x = 0.68375 x

x = 1.71 m

b) from figure 2

`F_1 = F_2`

`(kqQ_1)/(X^2) =(kqQ_2)/((2.88 + X)^2)`

`(Q_1)/(X^2) =(Q_2)/((2.88 +X)^2)`

FROM DATA GIVEN
`(7.85)/(x^2) = (3.67)/((2.88 + x)^2)`

solving for x we get

2.88 +x = 0.68375 x

x = - 9.1068 m