#1
here it is:

The electric potential energy, E, in joules, between two positively charged spheres of radius 10cm, separated by a distance, x, in meters, measured from the centres of the spheres, is given the eqn E=90/x. The force of repulsion, F, in newtons, on each shere is given by the rate of change of the electric potential with respect to x. Find the electric force of repulsion between the two spheres when they are separated by a distance of 2m.


WTF? id ont even knwo how to start this
#2
the key phrase in there is "given by the rate of change of the electric potential with respect to x"

that means differentiate that sucker. i assume you know how to do this...... if not, ill be back. when youve got that done, subs x=2 into the subsequent eqn, and thats it.
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#3
its more like physics, i guess it will need formulas from there and stuff...
sorry, i can't help, i quit physics last year in favor of biology (but i wouldn't have done so if it wasn't for the suckass teacher we had)
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#4



Most random lyrics ever...

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#5
first thts physics man not math and 2nd wat grade are u in? o and bout this well i dunno if this is of any help but the equation for tht is f=9*10^9 times q1 * q2 over r squared
#6
right, so i know i need to differience it (spelling) btu i dont knwo what exactly to sub in
plus. plsu the eqation form si ****ed not liek 3x2-3x=1 u know?
#7
right, so i know i need to differience it (spelling) btu i dont knwo what exactly to sub in
plus. plsu the eqation form si ****ed not liek 3x2-3x=1 u know?
**** differenciate...no id dont think we learnt that. you learn that next sem in calculus(new curriculum is ****ed) all i know is rate of change eqations
#9
how is this not a calculus problem? honestly........

Quote by rock_metal_fan
first thats physics man not math and 2nd wat grade are u in? o and bout this well i dunno if this is of any help but the equation for tht is f=9*10^9 times q1 * q2 over r squared

true, but its easier if done with calculus. yeah the general equation for this situation is 9*10^9*q1*q2/r^2, but that doesnt help him as he doesnt know what q1 or q2 are. and technically i think its supposed to be negative, but thats not that important when your figuring out the magnitude of the force. but i digress.

yeah this is a screwed up equation - here's how you make it less screwed up: it starts out as E=90/x. which is the same as saying E = 90*x^(-1). ya gotta differentiate that, and using the chain rule, you subtract 1 from the exponent on the x and make that its new exponent. then you multiply that new expression by the old exponent.

that should leave you with -90*x^(-2). which is the same as saying 90/(x^2). thats your equation that you substitute x=2 into to obtain 90/4 = 22.5N.
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#11
Quote by super666fender
right but we never learnt to differenciate. but what you say makes sense sorta

how does it not make sense? help me to clarify myself. although so long as you understand what's going on, thats all that matters.
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#14
Quote by super666fender
i dont understand because we never learnt to differentiate

i c what your sayin here, but i am unaware of the rate of change equations or whatever they are. which use calculus anyway....... i just did it the easiest way i could think of, which is using calculus.

oh well now you learned how to differentiate in a context-sensitive environment. the chain rule is a bit more complicated than what we used here, but only when you start getting parentheses raised to a power, etc.

you have a foot up on your calc class next semester as well.

id give you a tutorial of differentiation, but i only have so much space here, and thats 1/3 of what calculus is all about (the other 2/3 is integration = backwards differentiation, effectively).

hope that did it for ya
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#15
if its calculus then im in a ****ed up school cuz im takin precalculus and physics and its first bimester imagen how the **** im gonna know this, and im in 10 grade goddamit