Answer:
Please find the answer in the explanation
Explanation:
Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.
Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.
We can therefore conclude that the upper plate, is positively charged
B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer:
this measurement if feet is: 2.624672 ft
Explanation:
Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:
0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where


