A point charge of 5.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of –2.7 μC is placed placed on the x-axis at x2 = 0.23 m (Coulomb's law)?
A: Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
B: What if both charges are positive; that is, what if the second charge is 2.7 μC?
A: Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
B: What if both charges are positive; that is, what if the second charge is 2.7 μC?
1 Answer
(A) Let the third charge
Coulomb's Force on charge
F_(5.9->q)=k(5.9q)/x_3^2 .....(1)
Coulomb's Force on charge
F_(-2.7->q)=k(-2.7q)/(x_3-0.23)^2 ......(2)
For the net force to be zero, magnitude of both the forces must be equal.
|k(5.9q)/x_3^2|=|k(-2.7q)/(x_3-0.23)^2|
=>(5.9)/x_3^2=(2.7)/(x_3-0.23)^2
=>(5.9)(x_3-0.23)^2=(2.7)x_3^2
=>5.9x_3^2-2.714x_3+0.31211=2.7x_3^2
=>3.2x_3^2-2.714x_3+0.31211=0
Solving this quadratic with the help of in-built graphics tool we get my comp
two solutions as
x_3=0.14\ m and 0.71\ m
The direction of both forces must be opposite to each other.
(B) In such a case (2) becomes
F_(2.7->q)=k(2.7q)/(x_3-0.23)^2
The solutions for
Therefore, only valid solution would be