Solve 2.00 & 4.60µF Capacitors Voltage: Help Needed

  • Thread starter Thread starter ncsuwx
  • Start date Start date
  • Tags Tags
    Capacitance
AI Thread Summary
To find the voltage across the 2.00 µF and 4.60 µF capacitors when connected with opposite signs, the charge (Q) on each capacitor must be calculated using the formula C=Q/V. The total charge will redistribute when the capacitors are connected, affecting the final voltage. The user is struggling to apply the formula correctly and has noted a similar problem with different capacitor values. Assistance is requested to clarify the calculation process. Understanding the charge conservation principle is key to solving this problem effectively.
ncsuwx
Messages
3
Reaction score
0
Here is the question

A(n) 2.00 µF capacitor is charged to 520 V and a 4.60 µF capacitor is charged to 570 V

What is the voltage for each capacitor if plates of opposite signs are connected?

Any help would be helpful. I have bene trying to get this and haven't been able to all night. I only have one submission left to get the right answer soany suggestions would be greatly appreciated.
 
Physics news on Phys.org
The equation is C=Q/V. Just by pluggin in values does not work for me and i can't figure out what I am supposed to do.
 
I still can't figure it out. I have been trying and still can't get the right answer. I have an answer similar for a question for 2.4uF and 880V and 4uF and 560V capacitors. The voltage is 20V for that one. I can't even get that 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