Answer:
The distance between the particles have to be
x=0.01m or 1 cm
Explanation:
Using the law of Coulomb for two charge particles
[tex]F=K*\frac{q1*q2}{r^{2} } \\[/tex]
Notice you don't know the value of K or q1 or q2 however the value is the same in both cases, so you don't need to know exactly the value of each one you can put constant in the equation like this:
[tex]F=K*\frac{q1*q2}{r^{2} } \\[/tex]
[tex]50N= K*q1*q2*\frac{1}{(0.02m)^{2} } \\k*q1*q2=K[/tex]
[tex]K= 50N*(0.02m)^{2}\\ K=0.02[/tex] C
Now to find r can use the value of K
[tex]F=K*\frac{1}{r^{2} } \\r^{2}=\frac{K}{F} \\ r=\sqrt{\frac{0.02 C}{200N} } \\r=0.01m[/tex]
The distance r between the particles will be 1 cm.
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Electric charge consists of two types i.e. positively electric charge and negatively electric charge.
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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 !
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Given:
initial distance = d₁ = 2 cm
initial force = F₁ = 50 N
final force = F₂ = 200 N
Unknown:
final distance = d₂ = ?
Solution:
[tex]F_1 : F_2 = k \frac{q_1 q_2}{(d_1)^2} : k \frac{q_1 q_2}{(d_2)^2}[/tex]
[tex]F_1 : F_2 = \frac{1}{(d_1)^2} : \frac{1}{(d_2)^2}[/tex]
[tex]F_1 : F_2 = (d_2)^2 : (d_1)^2[/tex]
[tex]50 : 200 = (d_2)^2 : (2)^2[/tex]
[tex]1 : 4 = (d_2)^2 : 4[/tex]
[tex](d_2)^2 = 4 \div 4[/tex]
[tex]d_2 = 1 \texttt{ cm}[/tex]
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Grade: High School
Subject: Physics
Chapter: Static Electricity
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Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces
