The negative charge should be placed 41 m from +1.0 C charged object
Electric charge consists of two types i.e. positively electric charge and negatively electric charge.
There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let's tackle the problem now !
Given:
Q₁ = +1.0 C
Q₂ = +2.0 C
R₁ = x
R₂ = 100 - x
q = -1.0 × 10⁻³ C
Unknown:
x = ?
Solution:
In order to produce a total force = 0, the electric forces acting on this negative charge must be in the opposite direction and equal in magnitude.
We assume the distance of this negative charge to the positive charge +1.0 C is x, then:
[tex]F_1 = F_2[/tex]
[tex]k\frac{Q_1q}{R_1^2} = k\frac{Q_2q}{R_2^2}[/tex]
[tex]\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}[/tex]
[tex]\frac{1}{x^2} = \frac{2}{(100 - x)^2}[/tex]
[tex]\sqrt{\frac{1}{x^2}} = \sqrt{\frac{2}{(100 - x)^2}}[/tex]
[tex]\frac{1}{x} = \frac{\sqrt{2}}{100 - x}[/tex]
[tex]100 - x = \sqrt{2}x[/tex]
[tex]x + \sqrt{2}x = 100[/tex]
[tex]x ( 1 + \sqrt{2} ) = 100[/tex]
[tex]x = \frac{100}{1 + \sqrt{2}}[/tex]
[tex]x = 100(\sqrt{2} - 1) ~ m[/tex]
[tex]x \approx 41 ~ m[/tex]
Grade: High School
Subject: Physics
Chapter: Static Electricity
Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces
