Answer:

Explanation:
We can assume this problem as two concentric spherical metals with opposite charges.
We have also to take into account the formulas for the electric field and the capacitance. Hence we have

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating
![dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]](https://tex.z-dn.net/?f=dV%3DEdr%5C%5C%5C%5CV%3D%5Cint_%7BR_1%7D%5E%7BR_2%7DEdr%3D-%5Cint_%7BR_1%7D%5E%7BR_2%7D%5Cfrac%7BkQ%7D%7Br%5E2%7Ddr%5C%5C%5C%5CV%3DkQ%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D)
Hence, the capacitance is
![C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B1%7D%7Bk%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D%7D)
but R1=a and R2=b

HOPE THIS HELPS!!
Answer:
Therefore, the temperature at which the Fahrenheit scale reading is equal to half of the Celsius scale is −24.6∘C .
Area is calculated as length times width.
In this case, the width is 5 m and the length is 3 m.
5 • 3 = 15
So the area of Nicole's office is 15 square meters.
Square meters because only length and width is 2 dimensional.
Hope this helps!
Answer:
1) The medium b has higher optical density.
2) If the optical densities of both A and B are same.
Explanation:
- The refractive index of a material determines the number of times the velocity of light in that material is slower than the velocity of light in a vacuum.
- The higher refractive index means slower propagation of the light in that medium.
- The refractive index of the medium determines the optical density of the medium.
- If the material has a higher refractive index, then its optical density is high.
- If the material has a low refractive index, then its optical density is low.
- The light bends towards the normal while refracting through the medium from a lower optical density medium to the higher optical density medium.
- If both the mediums has the same optical density, the direction of propagation doesn't change.