Capacitor or EMW doubt
Consider a capacitor with UV light ionizing $n_{0}$ electrons per unit volume per unit second. Now each electron moves towards a plate. Given that plate 2 is at a higher potential than 1, and each electron moving ionizes $\alpha$ more electrons per unit length of the path find the number of electrons as a function of x and current density in the two plates.
20 Replies
SirLancelotDuLac
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Can also someone clarify the picture here? Shouldn't the number of electrons be a function of x and t?
(Given distance between plates is d)
Current density depends on time right
What's happening is UV rays are making the plate of capacitor emit photoelectrons (No per second) and they each ionise a more electrons as they move so no of electrons will be proportional (No)(t)(a)^x x is the distance traveled t time No the no of electrons emitted per second
With this you can calculate current dq/dt but I don't understand what to take as area to find current density
Lets just discuss the first part...
The uv light is ionizing the gas inside the capacitor right? So, if we have to find the exact function how would one go about it.
What I tried to do was take a cylindrical element of thickness dx (Assuming plates circular), and then I got lost :thinker:
Instead of cylindrical you could take thin cuboidal elements of thinkness dx then all the points on the cuboid would have to travel same distance to get to the other side
Saying cuboid would sound wrong since they are dx thickness it's almost like a plane
Ye like a sheet of dx thickness?
Will try this once.
Yep, could you share the answer.
The answer is 1 and 4
For the first part you can solve it buy writing the differential equation in terms of n(x)(no of electrons in that position)
d(n(x))[no of electrons ionised due to electrons travelling dx length]=adx(n(x))[adx tells how many electrons have been ionised due to n(x) &n(x) tells the no of electrons originally ionised by gas at that posn]
But what about the electrons generated due to the UV light?
n(x) takes them also into account
How so?
n(x) denotes all the electrons at that point and d(n(x)) denotes the electrons due to those points moving a dx length
The gas between the capacitor plates separated by a distance d is uniformly ionized by ultraviolet radiation so that $n{0}$ electrons per unit volume per second are formed. A constant potential difference is applied across the capacitor. These electrons moving in the electric field of the capacitor ionize the gas molecules each electrons producing $\alpha$ electrons per unit length of its path. Ionization by ions is neglected. Plate 1 is at a higher potential as compared to plate 2. Which of the following is correct?
1. The number of electrons per unit volume at a distance x from plate 2 is $n{0}e^{\alpha \cdot x}$
2. The number of electrons per unit volume at a distance x from plate 2 is $n{0}(1+\alpha \cdot x)$
3. The electronic current density at plate 1 is $en{0}e^{\alpha \cdot d}$
4. The electronic density at plate 1 is $\frac{en_{0}}{\alpha}(e^{\alpha \cdot d}-1)$
SirLancelotDuLac
is this a new q? start another thread na, people dont get pinged unless you do that
Well it is the same one but yeah its resolved.
+solved @hithav
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