Seperation effects on the forces of gravity and eletricity

AI Thread Summary
The discussion centers on the comparison between gravitational and electrical forces when two charged pellets are separated to twice their original distance. It concludes that the electrical force is significantly larger than the gravitational force due to the constants involved, with the gravitational force being "immensely smaller." When the distance is doubled, the electrical force decreases by a factor of four, while the gravitational force also decreases but remains negligible in comparison. The calculations demonstrate that both forces change by the same factor, but the electrical force remains dominant. Overall, the electrical force's strength and its greater change with distance highlight its significance in this context.
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Homework Statement


Two charged pellets are pulled apart to twice their original separation. (a) Which is likely to be larger, the gravitational force or the electrical force between them? Why? (b) Which will change by a greater factor when they are pulled apart, the gravitational force or the electrical force between them? Why?


Homework Equations


Fg = (G x m1 x m2) / r2
Fe = (K x q1 x q2) / r2


The Attempt at a Solution


a) Well, assuming 6.673 × 10−11 for the constant of G and 8.987 x 10 9 for the constant of K, I would assume that the electrical force would be higher.
b) I am assuming that they would change at the same factor when they are pulled apart.

Am I assuming correct or am I wrong?
 
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gravitational force is immmmmmmensely SMALLER than the electrical force.

remember, unless you are working with an extremely large mass, the electrical force will always be larger. plug in your value for mass and charge into the two force equations.
 
This type of question basically asks you what happens to the force when the distance is doubled. To see it, write
Fe = (K x q1 x q2) / r2
and then write it again replacing r with (2r). Do it with a red 2 if you can. Then you try to factor the 2 out so it sits in front of the rest of the stuff. In this case you'll get
Fe2 = (K x q1 x q2) / (2r)2 = 1/2^2*(K x q1 x q2) / r2 = 1/4*Fe
showing that the new force at double the distance is 1/4 of the original force.
 
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