- #1
jamie.j1989
- 79
- 0
Hi, I'm trying to measure the electric field of the Earth using a simple set up and would just like some opinions on the viability and practicality of this method.
The set up is as follows, two parallel metal plates with one larger than the other, the larger one is placed over the smaller at a separation of d, and an oscilloscope. I plan to ground the lower plate and to insulate the higher plate from ground. The larger plate should acquire a potential of,
$$V_{charge}=\frac{JA}{C}t$$
In a time t.
This result was found at http://arxiv.org/ftp/physics/papers/0701/0701296.pdf page 3 equation (4).
where J is the current density with units $$Amps*m^{-2}$$,A is the area of the plate, C is the capacitance, t the exposure time. I'm assuming here that what they mean by capacitance of the plate is the plates intrinsic capacitance? Which I am unsure about how to calculate.
I then plan to discharge the higher plate to the lower, and obtain a value for $$V_{charge}$$ from the oscilloscope, I will then use
$$\frac{V_{charge}C}{A}=\sigma$$
Where $$\sigma$$ is the charge density of the plate, and
$$\sigma=\epsilon_0E$$
To work out the electric field E of the earth. I'm slightly unsure whether the physics is correct though?
Thanks for any help.
The set up is as follows, two parallel metal plates with one larger than the other, the larger one is placed over the smaller at a separation of d, and an oscilloscope. I plan to ground the lower plate and to insulate the higher plate from ground. The larger plate should acquire a potential of,
$$V_{charge}=\frac{JA}{C}t$$
In a time t.
This result was found at http://arxiv.org/ftp/physics/papers/0701/0701296.pdf page 3 equation (4).
where J is the current density with units $$Amps*m^{-2}$$,A is the area of the plate, C is the capacitance, t the exposure time. I'm assuming here that what they mean by capacitance of the plate is the plates intrinsic capacitance? Which I am unsure about how to calculate.
I then plan to discharge the higher plate to the lower, and obtain a value for $$V_{charge}$$ from the oscilloscope, I will then use
$$\frac{V_{charge}C}{A}=\sigma$$
Where $$\sigma$$ is the charge density of the plate, and
$$\sigma=\epsilon_0E$$
To work out the electric field E of the earth. I'm slightly unsure whether the physics is correct though?
Thanks for any help.