Electrostatics - metal sphere potential

AI Thread Summary
A metal sphere of radius R1 is initially charged to 10V and then connected to a larger sphere with radius R2 equal to 2*R1. The final potential of the connected spheres is determined by distributing the charge so that their potentials are equal. The calculations show that the potential of the connected spheres is V_connected = V/2, which simplifies to 5V. This conclusion aligns with the principle that when two conductors are connected, they share charge until their potentials equalize.
xwhyy
Messages
5
Reaction score
0
Hi,

again very simple problem

Homework Statement



Metal sphere of radius R1 is charged to 10V.
It is then connected to another sphere with R2 = 2*R1.
What is the final potential?

Homework Equations





The Attempt at a Solution



V = E * distance
so in the case of disconnected it is E*R1
when connected R's sum up so it would be E*(R1+R2)
and since I don't know E but R2=2*R1
it is E*3*R1 so my solution is:

V_connected = V * 3

Is that correct?
 
Physics news on Phys.org
If it's correct, that's only a coincidence.

What is the potential of a sphere of radius R1 and charge Q?

Set that equal to 10V and solve for Q.

Then distribute the charge over the two spheres so that their potentials are equal.
 
Thank you for your reply!

so:

V = Q / [4*pi*epsilon*r1]
Q = V * 4* pi * epsilon * r1

V_connected = (V * 4 * pi * epsilon * r1) / ( 4 *pi * epsilon * 2 * r1)
which after simplification is equal
V_connected = V/2

is that true? :)
 
hop hop
 
xwhyy said:
...

V = Q / [4*pi*epsilon*r1]
Q = V * 4* pi * epsilon * r1
Looks good. BTW: V = 10 Volts

When they're connected: V1 = V2 and Q1 + Q2 = Q = V·4πε0·R1.
 
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