Recent content by xNick94

Sorry for the late reply, had to work and family friends were visiting. Would you then have to multiply it since 4/9 is under 1. 3.1x10^-8 x 4/9 = 1.3 x 10^-8 Would the force be just 1.3x10^-8 or would i have to subtract that from 3.1x10^-8?

Hmm so was my original divide procedure justified from my explanation? 3.1x10^-8 / (4/9) = 6.9x10^-8 ...the force increased so this has to be wrong since it has to decrease right?

Yes good idea, i'll stick with 4/9 instead of the decimal to preserve accuracy. Since F is proportional to 1/R^2. If R increases from 3 m to 4.5 m shouldn't the force decrease by a factor of (3/4.5)^2. And i thought decrease by a factor means divided by so i divide the original force by the...

Alright that makes sense, thanks for all your help!

I'm really confused :( I'm trying to understand it but I'm having trouble grasping the concept - F is proportional to R^2 since the R^2 came from F=kQq/R^2. So why doesn't that make it an inversely proportional graph or what is it even supposed to be- this is really frustrating :(

Two charges Q1 amd Q2 jabe a force of 3.1x10^-8 N when the distance between them is 3m. Use the formula F = kQq/R2 to determine the force when the distance (R) between charge Q1 and Q2 increased from 3m to 4.5m The Attempt at a Solution My attempt was that F = 1/R^2 k= 9.0x10^9...

Thanks for your reply, I suspected it could be Square.root 1/R. and does it really make a difference if you express it in the other way?

My first post, yay i already like the atmosphere here :P anyway... Using the formula F = kQq/R2 sketch graphs between a. F and Q (k,q, and R are constant) b. F and R (Q,q and k are constant) c. Q and R (k,q and F are constant) I think i did it correctly but I'm not quite sure...