contestada

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?

Respuesta :

Answer:

x = 1.71 m

x = - 9.1068 m

Explanation:

from figure 1

[/tex]f_1 = f_2[/tex]

[tex]\frac{kqQ_1}{X^2} =\frac{kqQ_2}{(2.88-X)^2}[/tex]

[tex]\frac{Q_1}{X^2} =\frac{Q_2}{(2.88-X)^2}[/tex]

FROM DATA GIVEN

[tex]\frac{7.85}{x^2} = \frac{3.67}{(2.88-x)^2}[/tex]

solving for x we get

2.88 -x = 0.68375 x

x = 1.71 m

b) from figure 2

[tex]F_1 = F_2[/tex]

[tex]\frac{kqQ_1}{X^2} =\frac{kqQ_2}{(2.88 + X)^2}[/tex]

[tex]\frac{Q_1}{X^2} =\frac{Q_2}{(2.88 +X)^2}[/tex]

FROM DATA GIVEN [tex]\frac{7.85}{x^2} = \frac{3.67}{(2.88 + x)^2}[/tex]

solving for x we get

2.88 +x = 0.68375 x

x = - 9.1068 m

Ver imagen rejkjavik
Ver imagen rejkjavik

Otras preguntas