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Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 1 is doubled AND the distance separating Objects 1 and 2 is tripled, then the new electrostatic force will be _____ units.

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skyluke89
The electrostatic force between two charged objects is given by
[tex]F=k \frac{q_1 q_2}{r^2} [/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

In this problem, the charge of object 1 is doubled:
[tex]q_1' = 2 q_1[/tex]
while the distance between the two objects is tripled:
[tex]r'=3 r[/tex]
Therefore, the new electrostatic force will be
[tex]F'=k \frac{q_1' q_2}{(r')^2}=k \frac{2 q_1 q_2}{(3r)^2} = \frac{2}{9}(k \frac{q_1 q_2}{r^2}) = \frac{2}{9} F [/tex]

And since the original force was F=18.0 units, the new force is
[tex]F= \frac{2}{9} (18.0 u)=4 u [/tex]
shubhamchouhanvrVT

If the charge of Object 1 is doubled and the distance separating Objects 1 and 2 is tripled then-new the electrostatic force will be 4 units.

What will be the new electrostatic force?

It is given that

The force between charge 1 and charge 2 [tex]F_1=18\ umits[/tex]

From Columb's law of electrostatic forces

[tex]F_1=\dfrac{kq_1 q_2}{r_1^2}[/tex]

Now from the given condition

A New charge1 will be [tex]q_1'=2q_1[/tex]

The new radius will be [tex]r'_1=3r_1[/tex]

and the charge [tex]q_2[/tex] will remain constant

Now put the new values in the formula

[tex]F_2=\dfrac{kq'_1 q'_2}{r'_1^2}[/tex]

[tex]F_2=\dfrac{2}{9} \dfrac{kq_1 q_2}{r_1^2}[/tex]

[tex]F_2=\dfrac{2}{9} F_1[/tex]

Since

[tex]F_1=\dfrac{kq_1 q_2}{r_1^2} =18\ units[/tex]

[tex]F_2=\dfrac{2\times 18}{9} =4[/tex]

Thus If the charge of Object 1 is doubled and the distance separating Objects 1 and 2 is tripled then-new the electrostatic force will be 4 units.

To know more about electrostatic force follow

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