- #1
kostoglotov
- 234
- 6
Homework Statement
The Question is here:
imgur link: http://i.imgur.com/wIlUZMM.jpg
Homework Equations
[tex]U_c = \frac{\kappa \epsilon_0 A d E^2}{2}[/tex]
[tex]U_c = C \Delta V_c[/tex]
[itex]\kappa[/itex]: dielectric constant
[itex]\epsilon_0[/itex]: permittivity of free space
[itex]C[/itex]: capacitance
The Attempt at a Solution
[/B]
I understand why the answer to (a) is no, conservation of mass, and the charge can't flow anywhere since the system has been isolated.
But I got completely the opposite answers to (b) and (c) from a solution I found posted online.
My answer:
(b) The electric field strength decreases by a factor of [itex]\frac{1}{\sqrt{2}}[/itex]. This is required by the conservation of energy and the equation [itex]U_c = \frac{\kappa \epsilon_0 A d E^2}{2}[/itex]. Since the [itex]U_c[/itex] cannot change, nor can the [itex]\frac{\kappa \epsilon_0 A}{2}[/itex] all that [itex]U_c[/itex] is stored in the electric field. After all, a certain amount of work had to be done on the capacitor by a source to charge it to a particular voltage, and that work/energy can't just disappear. Will increased separation there is now more space for the electric field to "change the shape of", so now any given set of equipotential surfaces are now more spread out, lower V per meter, but more distance, lower force per Coulomb, less force on a given charge, but over a greater distance to give an overall same [itex]\Delta K[/itex].
(c) No, [itex]\Delta V_c[/itex] stays the same as it is proportional to the work done by the original energy source in separating the charges across the capacitor, so conservation of energy would prevent [itex]\Delta V_c[/itex] from changing.
The answer in a solution I found online is:
imgur link: http://i.imgur.com/SMSdxiO.jpg
The problem is, these solutions don't explain very much conceptually.
Firstly, am I wrong? Secondly, why? Why is the answer the opposite of what I've answered?