Answer:
Explanation:
Let the third charge Qc = + 3 μC be placed at distance x from origin so that potential at origin due to all charges be zero .
potential due to charge Q at distance d
= k Q / d , k = 9 x 10⁹
so potential due to all charges at origin
9 x 10⁹ x 10⁻⁶ [tex](\frac{2}{.01} -\frac{6}{.02} + \frac{3}{x} )[/tex] = 0
[tex]\frac{3}{x}[/tex] = [tex]-\frac{2}{.01}[/tex] [tex]+\frac{6}{.02}[/tex]
= - 200 + 300
= 100
x = .03 m
3 cm .
+ 3μC should be placed at + 3 cm from origin.
3μC should be placed at + 3 cm on the positive axis from the start
From the question,
the formula for the potential charge is k Q / d , k
where q is charge
d is distance
k is constant.
let the third charge be Qc = + 3 μC at distance x from origin so that potential at origin due to all charges be zero .
potential of charge Q at distance d
= k Q / d , k = 9 x 10⁹
potential of all charges at zero is
9 x 10⁹ x 10⁻⁶ = 0{-2/0.01 -6/0.02 + 3/x = 0
3/x= -2/0.01 + 6/0.01
= - 200 + 300
= 100
x = .03 m
3 cm .
Electric charge is the property found in matter that make it to experience a force when it is place in an electromagnetic field. Electric charge can either be positive or negative. It is represented by q.
Therefore, + 3μC should be placed at + 3 cm from the start.
Learn more about charge from the link below.
https://brainly.com/question/18102056
