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.

Guest Apr 10, 2015

#2**+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

#1**+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**+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