1sir,
seems dat no one is tryin dis prob.........
pls guide me d ans before d encounter exam
A plane divides space into two halves. One half is filled with a homogeneous conducting medium and physicists work in the other. They mark the outline of a square of side a on the plane and let a current I_0 in and out at two of its neighboring corners using fine electrodes. Meanwhile, they measure the potential difference V between the other two corners. (see figure)
What is the resistivity of the homogeneous medium in terms of the given data?
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32 Answers
1gr8 solution sir..........
but still not gettin it completely..
ve to look at d solution more carefully..........
66Okay, here is the solution. Usually, when encountering problems having infinite extensions, superposition principle comes to mind. Here also, we use superposition to combine separate discussions of current flowing in and out. First, we send the current into A and collect it at infinity. Then, we send the current at infinity and collect it at B. When we superpose the two situation, we get the situation of the problem.
When the current I0 is sent into A and collected at infinity, it is distributed hemispherically symmetrically in the half plane containing the conducting medium. That means that the current density j at a distance r from A is given by
j(r)=\dfrac{I_0}{2\pi r^2}
As per Ohm's law in the local form \vec{j}(r)=\dfrac{\vec{E}(r)}{\rho} (here \rho is the resistivity and \vec{E}), we get the magnitude of electric field at a distance r from A as
E(r)=\dfrac{I_0\rho}{2\pi r^2}
The potential at a distance r can be obtained by integrating the field. However, in this case, we get it by comparison of a point charge. We note that the field of a point charge varies as \dfrac{1}{r^2} and the potential only as \dfrac{1}{r} and the coefficient of \dfrac{1}{r^2} and \dfrac{1}{r} are the same. In the present case also the field is varying as \dfrac{1}{r^2}, so the potential must be
V(r)=\dfrac{I_0\rho}{2\pi r}
Hence, the potential difference between C and D, when current is sent at A and collected at infinity is
\Delta V_1=\dfrac{I_0\rho}{2\pi a}-\dfrac{I_0\rho}{2\pi a\sqrt{2}}=\dfrac{I_0\rho}{4\pi a}(2-\sqrt{2})
We next send the current at infinity and collect it as B. Everything is the same except the signs of the quantities. The potential difference between C and D is therefore the same.
So when we superpose, the potential difference between C and D becomes twice of what we calculated above. Hence,
V_0=2\Delta V_1=\dfrac{\rho I_0}{2\pi a}(2-\sqrt{2})
Accordingly, the resistivity can be found as
\boxed{\rho = (2+\sqrt{2})\dfrac{\pi a V_0}{I_0}}
11resistivity can be given as rho=E/J
E-ELECTRIC FIELD
J-CURRENT DENSITY...
i think m makin a mistake in calculatin d field........... cldn get d question properly.........
66Find the electric field inside the conducting half plane.. that's the hint [1]
1U ASKED ME WAT IS E ......... E - electric field.....
giv me a toughest clue to find d ans........;-)
1sir...................................................;-)
1u ve mistaken............. sir is askin resistivity,,,, not resistance.....