Answer: It will remain the same
Explanation:
According to Coulomb's Law:
"The electrostatic force [tex]F_{E}[/tex] between two point charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is proportional to the product of the charges and inversely proportional to the square of the distance [tex]d[/tex] that separates them, and has the direction of the line that joins them".
Mathematically this law is written as:
[tex]F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}[/tex] (1)
Where:
[tex]F_{E}[/tex] is the electrostatic force
[tex]K[/tex] is the Coulomb's constant
[tex]q_{1}=q_{2}=q[/tex] are the electric charges , which in this case have the same positive charge
[tex]d[/tex] is the separation distance between the charges
Rewritting we have:
[tex]F_{E}=K\frac{q^{2}}{d^{2}}[/tex] (2)
Now, if the first charge is doubled:
[tex]q_{1}=2q_[/tex]
And the second is reduced to a half:
[tex]q_{2}=\frac{1}{2}q[/tex]
We will have the following:
[tex]F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}}[/tex] (3)
[tex]F_{E}=K\frac{q^{2}}{d^{2}}[/tex] (4)
As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same
Answer:
The Answer Is A my evidence is the formula above
Explanation:
