Two positively charged spheres, each with a charge of 2.0 *
10-5 ...

Question

Two positively charged spheres, each with a charge of 2.0 *
10-5 C, a mass of 1.0 kg, and separated by a distance of
1.0 cm, are held in place on a frictionless track. (a) What
is the electrostatic potential energy of this system?

Accepted Answer

Hi everyone for this problem, we're told to consider two spheres that are separated by a distance of 3. cm and are held in place on a frictionless track. The first fear has a charge of 3.5 times 10 to the negative six column and a mass of 1.5 kg. The second sphere has a charge of 4.2 times 10 to the negative six Kalume and a mass of 1.2 kg were asked to calculate the electrostatic potential energy of the system. And so our equation that will allow us to do that is our electrostatic potential energy is equal to K times Q one times Q two over Our distance. RK represents columns constant and I'll make a note of that on the side and columns constant is 8.99 times 10 to the ninth jewel meters over column squared. So this is something that we should know. It wasn't given in the problem Q one and Q to represent our charges. And they told us this in the problem, they told us that the first fear has a charge of 3.5 times 10 to the negative six Kalume. And our second charge has our second sphere has a charge of 4.2 times 10 to the negative six column. So we have both of those. We have Q one and Q two. Okay. And lastly our d equals r distance. They tell us in the problem that the two spheres are separated by a distance of 3.5 cm. So we're going to need to convert this to meters because our our unit in our K constant is meters. So in one m we have 100 centimeters so are Conversion two m is going to equal 0.035 m. Okay, so now we have everything that we need. All we have to do is plug in and solve for E. R. Electrostatic potential energy. So let's go ahead and do that. So our electrostatic potential energy Is going to equal 8.99 times 10 to the 9th jewel meters over column squared times both of our charges. And this is all over our distance squared. All over our distance. It's not squared. Excuse me. Okay, so once we calculate this out, we'll get a final answer of 3.8 jewels. This is our electrostatic potential energy of the system. And that is the end of this problem. I hope this was helpful.

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