Answer:
![\rm 9.186\times 10^{-7}\ C.](https://tex.z-dn.net/?f=%5Crm%209.186%5Ctimes%2010%5E%7B-7%7D%5C%20C.)
Explanation:
<u>Given:</u>
- Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
- Distance of separation between the plates, d = 1.0 cm = 0.01 m.
- Minimum value of electric field that produces spark,
![\rm E=3\times 10^6\ N/C.](https://tex.z-dn.net/?f=%5Crm%20E%3D3%5Ctimes%2010%5E6%5C%20N%2FC.)
When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by
![\rm E=\dfrac{\sigma}{\epsilon_o}.](https://tex.z-dn.net/?f=%5Crm%20E%3D%5Cdfrac%7B%5Csigma%7D%7B%5Cepsilon_o%7D.)
where,
= surface charge density of the plate of the capacitor =
.
= magnitude of the charge on each of the plate.
= surface area of each of the plate =![\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.](https://tex.z-dn.net/?f=%5Crm%20%5Cpi%20%5Ctimes%20%28Radius%29%5E2%3D%5Cpi%20%5Ctimes%5Cleft%20%28%20%5Cdfrac%7BD%7D%7B2%7D%5Cright%20%29%5E2%3D%20%5Cpi%20%5Ctimes%20%5Cleft%20%28%20%5Cdfrac%7B0.21%7D%7B2%7D%5Cright%20%29%5E2%3D3.46%5Ctimes%2010%5E%7B-2%7D%5C%20m%5E2.)
= electrical permittivity of free space, having value = ![8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.](https://tex.z-dn.net/?f=8.85%5Ctimes%2010%5E%7B-12%7D%5Crm%20%5C%20C%5E2N%5E%7B-1%7Dm%5E%7B-2%7D.)
For the minimum value of electric field that produces spark,
![\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.](https://tex.z-dn.net/?f=%5Crm%20E%20%3D%20%5Cdfrac%7Bq%7D%7BA%5Cepsilon_o%7D%5C%5C%5CRightarrow%20q%20%3D%20E%5C%20A%5Cepsilon_o%5C%5C%3D3%5Ctimes%2010%5E6%5Ctimes%203.46%5Ctimes%2010%5E%7B-2%7D%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5C%5C%3D9.186%5Ctimes%2010%5E%7B-7%7D%5C%20C.)
It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.
Answer:
Speed is the rate at which an object's position changes, measured in meters per second. The equation for speed is simple: distance divided by time
Explanation:
Answer:
65
Explanation:
The resonant frequencies for a fixed string is given by the formula nv/(2L).
Where n is the multiple
.
v is speed in m/s
.
The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn
fundamental frequency means n=1
i.e fn+1 – fn = 390 -325
= 65
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula ![E = \dfrac{kq }{d}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bkq%20%7D%7Bd%7D)
![E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B9%20%5Ctimes%2010%5E9%20%5Ctimes%20%2833.3%20%5Ctimes%2010%5E%7B-9%7D%29%20%7D%7B0.205%7D)
E = 1461.95 N/C
c) The electric field E is calculated as:
![E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B9%20%5Ctimes%2010%5E9%20%5Ctimes%20%2833.3%20%5Ctimes%2010%5E%7B-9%7D%29%20%7D%7B1.25%7D)
E = 239.76 N/C