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At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is tripled, then the repulsive force would be ____ N.

physics
Guest Apr 10, 2015

Best Answer 

 #2
avatar+91451 
+5

Hi Alan,

I was just thinking about this.

If you put one of the balloons at the centre of a sphere with a radius of r units.  

Then for that sphere  the       $$Surface \;area=4\pi r^2$$

The attractive force will be the same for every point on the surface of that sphere.

Now

It is a concentic sphere encases the first one and it has a radius of  3r then

For this second sphere    

 

   $$\\Surface \;area=4\pi (3r)^2=4\pi*9r^2=9*(4\pi r^2) \\= 9*$ the original spheres surface area$\\\\$$

 

So the force would have to be 9 times less if the radius is 3 times more.  

 

Does that make sense Alan?

Melody  Apr 11, 2015
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2+0 Answers

 #1
avatar+26402 
+5

Electrostatic force is inversely proportional to the square of the distance between the two balloons, so if this distance is tripled the force will be 9 times smaller, namely:   0.32/9 N

.

Alan  Apr 10, 2015
 #2
avatar+91451 
+5
Best Answer

Hi Alan,

I was just thinking about this.

If you put one of the balloons at the centre of a sphere with a radius of r units.  

Then for that sphere  the       $$Surface \;area=4\pi r^2$$

The attractive force will be the same for every point on the surface of that sphere.

Now

It is a concentic sphere encases the first one and it has a radius of  3r then

For this second sphere    

 

   $$\\Surface \;area=4\pi (3r)^2=4\pi*9r^2=9*(4\pi r^2) \\= 9*$ the original spheres surface area$\\\\$$

 

So the force would have to be 9 times less if the radius is 3 times more.  

 

Does that make sense Alan?

Melody  Apr 11, 2015

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