Electrostatics and potential question

AI Thread Summary
To calculate the work done on a 2.5C charge moved from a potential of 12V to 75V, first find the voltage difference, which is 63V. The work done can be determined by multiplying this voltage difference by the charge, using the formula W = QΔV. It's crucial to maintain unit consistency throughout the calculation, as 1 Volt equals 1 Joule per Coulomb. Attention to units is emphasized as essential for understanding the physics and achieving accurate results. Properly applying these principles will lead to confidence in solving similar problems.
Inertialforce
Messages
67
Reaction score
2

Homework Statement


A 2.5C charge is moved from a point with a potential of 12V to another point of potential 75V. How much work was done on this charge?


Homework Equations


ΔV = ΔEp/Q


The Attempt at a Solution


I am unsure how to do this question but what I did do is I took the two potentials and I subtracted the two to get a difference in potential. Then I took that difference in potential and divided it by the charge of 2.5C to get the change in voltage.

Now that I have the change in voltage what equation am I supposed to use to find the amount of work done?
 
Physics news on Phys.org
The change in electric potential is the change in voltage. Your equation would better be written Q\Delta V = \Delta E. What is the p in your equation?
 
jambaugh said:
The change in electric potential is the change in voltage. Your equation would better be written Q\Delta V = \Delta E. What is the p in your equation?

Are you talking about the "p" in ΔEp? If so then the "p" together with the "E" stands for electric potential (because that's how it is written in the formula sheet that is given to us).

So I take the two point potentials and find the difference between them and multiply that by the charge to get ΔEp (I tried that and got a ΔEp= 63)? Then what would I do with that ΔEp?
 
Inertialforce said:
So I take the two point potentials and find the difference between them and multiply that by the charge to get ΔEp (I tried that and got a ΔEp= 63)? Then what would I do with that ΔEp?
What do you understand by 'ΔEp' in your given equation?
 
I don't know who wrote your formula sheet but it appears to be wrong. If you define \Delta E as the change in energy and \Delta V as the difference in voltage (electric potential) and drop the p then your formula would be right.

Look at the units, look at the units, look at the units!

1 Volt = 1 Joule per Coulomb. You have the formula built into the units.
Volts times Coulombs equals Joules
Voltage difference times charge equals energy!

w.r.t. 63 check your arithmetic and keep your units. The difference in potentials is 75volts - 12volts = 63volts. Multiply that by the charge to get the energy.

I'll say it some more... units units units units units units... and I can't emphasize enough UNITS! Raw numbers are meaningless. If you want to understand the physics you MUST pay attention to units. If you want to do well on tests you really need to understand the physics.

Sorry to vent at you but this is one of my biggest gripes with my students. They drop units and loose any track of what they're doing. Remember that you aren't saving time or work if you have to repeat the attempts over and over. Work it once with the units carefully included and you'll get there more quickly and with, what is more important, confidence in your answer.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top