An object with a charge of +28 µC is placed at the origin. Another object with a charge of +49 µC is placed a distance of +36 cm from the first object along the x axis. A third object with charge -17 µC is moved from the (x,y) location (0,10) to (12,10), where the positions are given in cm. When the third object is moved this distance, what is the change in the potential energy?

Question

anonymous · Answer

You know the equation for electrostatic potential due to  a point charge ?

anonymous · Answer

kQ/r

anonymous · Answer

ok, so if  two charges Q and q are interacting through the electrostatic potential, what is their mutual potential energy ?

anonymous · Answer

Add them?

anonymous · Answer

No

anonymous · Answer

what is the electrostatic potential energy of charges Q and q with a separation r between them ?

anonymous · Answer

I dont know what you are asking

anonymous · Answer

You gave me a formula above, kQ/r is what you typed
Do you know what the physical significance of that formula is ?

anonymous · Answer

Yes

anonymous · Answer

Tell me

anonymous · Answer

Electric potential energy?

anonymous · Answer

The electric potential energy of what ?

anonymous · Answer

I don't know

anonymous · Answer

So you have learnt something already - being able to write an equation down, or knowing the name of an equation, is not the same as understanding how to apply that equation in problems

anonymous · Answer

OK. Can you help me solve this problem?

anonymous · Answer

yes, that's what i am doing, by helping you to understand what potential energy is

anonymous · Answer

Ok. But that doesn't help right now.

anonymous · Answer

well if you just want me to give you the answer you won't really learn anything

anonymous · Answer

I can learn after.

anonymous · Answer

i will get you started
the potential energy of two point charges Q and q, separated by a distance r, is
kQq/r

anonymous · Answer

Now that is all you need to know to solve your problem

anonymous · Answer

I already tried it. I keep getting the wrong answer

anonymous · Answer

almost all
you also need to assume that the potential energy of several charges is given by the sum of the potential energies of the separate pairs of charges calculated according to the law given above

anonymous · Answer

ah ok, well should we go through some of your numbers ?

anonymous · Answer

3.81e-9 for q1q2

-4.76e-9 for q1q3

-2.25e-9 for q2q3

anonymous · Answer

Then after moving

3.81e-9 for q1q2

-3.05e-9 for q1q3

-3.2e-9 for q2q3

anonymous · Answer

so these are the potential energies between the pairs of charges ?

anonymous · Answer

yes

anonymous · Answer

I add them up and then subtracted the difference

anonymous · Answer

Can you tell me what you get?

anonymous · Answer

what are your units of energy ?

anonymous · Answer

Joules

anonymous · Answer

by the way, you don't need to include the potential energy of the 28 and 49 charges because that doesn't change, you can ignore it for calculating differences

anonymous · Answer

you must have gone wrong somewhere, my numbers are much larger

anonymous · Answer

let's break it down further to find out where you have gone wrong

anonymous · Answer

What did you get?

anonymous · Answer

looking at the initial situation, before the -17 charge moves, what do you have for the separation between the -17 and the 49 charges

anonymous · Answer

.37m

anonymous · Answer

yes, same here more or less, 37.36 cm

anonymous · Answer

so their potential energy is $$-k*17e-6*49e-6/0.3736$$

anonymous · Answer

what is your value of k ?

anonymous · Answer

9e9

anonymous · Answer

so what do you get for that energy above if you work out the numbers ?

anonymous · Answer

Like 6.84J

anonymous · Answer

no, my answer is different

anonymous · Answer

What did you get?

anonymous · Answer

i don't want to just give u the numbers, i want you to get them yourself

anonymous · Answer

It would help me know where I went wrong

anonymous · Answer

well all you have to do is calculate 9e9 x 17e-6 x 49e-6 and divide it by 37.36

anonymous · Answer

try it again and tell me what you get

anonymous · Answer

sorry, divide by 0.3736, it should be in meters

anonymous · Answer

20.1J?

anonymous · Answer

yes

anonymous · Answer

Is that the answer?

anonymous · Answer

now you need to consider the initial situation again and calculate the energy between -17 and 28

anonymous · Answer

what is their separation to start with ?

anonymous · Answer

.1

anonymous · Answer

that's right

anonymous · Answer

now u can figure out their potential energy before the move

anonymous · Answer

42.84 + 20.1?

anonymous · Answer

yes, but remember they are both negative energies

anonymous · Answer

ok

For after the move I got -27.5 and -28.8

anonymous · Answer

so to start with we have -62.9 joules

anonymous · Answer

well if we look at the arrangement after the move, what separations do you have ?

anonymous · Answer

.156 and .26

anonymous · Answer

yes that's what i have

anonymous · Answer

Ok So do I add or subtract the values?

anonymous · Answer

but your second value is not correct

anonymous · Answer

-28.8?

anonymous · Answer

oops sorry, i was looking at the wrong number, your second number is correct

anonymous · Answer

so initial energy was -62.9J and final energy was -56.26J

anonymous · Answer

so Vfinal - Vinitial = ?

anonymous · Answer

6.64?

anonymous · Answer

I think so

anonymous · Answer

Ok thanks. Must have made some technical mistake. Initially got 6.84J

anonymous · Answer

Were you given the correct answer ? do you know what the answer is supposed to be ?

anonymous · Answer

did we get it right ?

anonymous · Answer

yes, no, don't care . . . . .