- #1
issacnewton
- 1,000
- 29
Homework Statement
Hi
This is a problem from problem book by russian author, Igor Irodov.
Four large metal plates are located at a small distance d from one another as shown in the figure. The extreme plates are connected by a metal wire, while the potential difference V is applied to internal plates. Find
a)the values of electric field strength between neighboring plates
b) the surface charge density on each plate.
Homework Equations
[tex]V_a - V_b= \int_a^b \overrightarrow{E}\circ \overrightarrow{dl}[/tex]
The Attempt at a Solution
I think I got the solution since my answer matched with the answer given at the back of the book. But I have some questions.
First I imagined just plates 2 and 3 connected to a battery. Now battery makes sure that the
two plates will remain at a given potential difference. So now if we bring plates 1 and 4
close to 2 and 3 (which is basically a capacitor), the potential difference between 2 and 3 will not change. And since the problem mentions that 2 and 3 have potential difference of V in the new configuration, the potential difference between 2 and 3 was V in the beginning , before we decided to bring the plates 1 and 4 close to them. Now here I am assuming that as we bring 1 and 4 closer to 2 and 3, the capacitance of 2 and 3 also does not change. I don't know if this assumption is correct though. So with this assumption, since [itex]C=q/V[/itex], and since C and V are constants, charge q will also not change as bring plates 1 and 4 closer.
Now by the symmetry of the problem, E between 1 and 2 will be same as E between 3 and 4. Let's call this E1. Let E between 2 and 3 be E2. Now since 1 and 4 are at the same potential, if point a is on 1 and point b is on 4, then [itex]V_a - V_b =0[/itex]. But
[tex]V_a - V_b= \int_a^b \overrightarrow{E}\circ \overrightarrow{dl}[/tex]
where path goes through 2 and 3 and its straight. So we get
[tex]0=-E_1 d+V -E_1 d[/tex]
which means [itex]E_1=\frac{V}{2d}[/itex]. Also [itex]V= E_2 d[/itex], so we get
[tex]E_1=\frac{V}{2d},\;\;E_2= \frac{V}{d}[/tex]
Now let [itex]\sigma_1 , \sigma_2[/itex] be the magnitudes of the surface charge densities
on plate 1 and plate 2. Using the fact that electric field due to surface charge density [itex]\sigma[/itex] on a metal plate is given by [itex]E=\frac{\sigma}{2\epsilon_o}[/itex]
on both the sides of the plane, I derived the surface charge densities on plates. I got
[tex]\sigma_1=\frac{\epsilon_o V}{2d},\;\;\sigma_2=\frac{3 \epsilon_o V}{2d}[/tex]
Ok, my answers are correct. But I have question. I made the assumption that as we bring plates 1 and 4 closer to 2 and 3, the capacitance of 2 and 3 remains constant. Is that valid ? and why ? Wont bringing some conductors closer to some capacitor change its capacitance ? Because if we insert a dielectric between the capacitor plates, its capacitor changes.
So it may happen in the situation of this problem.
thanks