Answer:
The permittivity of rubber is ![\epsilon = 8.703 *10^{-11}](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%208.703%20%2A10%5E%7B-11%7D)
Explanation:
From the question we are told that
The magnitude of the point charge is ![q_1 = 70 \ nC = 70 *10^{-9} \ C](https://tex.z-dn.net/?f=q_1%20%3D%20%2070%20%5C%20nC%20%20%3D%20%2070%20%2A10%5E%7B-9%7D%20%5C%20%20C)
The diameter of the rubber shell is ![d = 32 \ cm = 0.32 \ m](https://tex.z-dn.net/?f=d%20%3D%2032%20%5C%20cm%20%20%3D%20%200.32%20%5C%20m)
The Electric field inside the rubber shell is ![E = 2500 \ N/ C](https://tex.z-dn.net/?f=E%20%3D%20%202500%20%5C%20N%2F%20C)
The radius of the rubber is mathematically evaluated as
![r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m](https://tex.z-dn.net/?f=r%20%3D%20%20%5Cfrac%7Bd%7D%7B2%7D%20%3D%20%20%5Cfrac%7B0.32%7D%7B2%7D%20%20%3D%20%200.16%20%5C%20m)
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
Where
is the permittivity of rubber
=> ![E * \epsilon * 4 * \pi * r^2 = Q](https://tex.z-dn.net/?f=E%20%20%2A%20%20%5Cepsilon%20%20%2A%20%204%20%2A%20%5Cpi%20%2A%20%20r%5E2%20%3D%20%20Q)
=> ![\epsilon = \frac{Q}{E * 4 * \pi * r^2}](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%20%20%5Cfrac%7BQ%7D%7BE%20%2A%20%204%20%2A%20%20%5Cpi%20%2A%20%20r%5E2%7D)
substituting values
![\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%20%20%5Cfrac%7B70%20%2A10%5E%7B-9%7D%7D%7B2500%20%2A%20%204%20%2A%20%203.142%20%2A%20%20%280.16%29%5E2%7D)
![\epsilon = 8.703 *10^{-11}](https://tex.z-dn.net/?f=%5Cepsilon%20%20%3D%208.703%20%2A10%5E%7B-11%7D)
Constant speed along the inclined surface = 30 m / 4 s = 7.5 m/s
Vertical speed = inclined speed * sin(45) = 7.5 *√2 / 2 = 5.3 m/s
Answer: 5.3 m/s
1. 2 way radio
2. radio waves
3. communication
4. convert the voltage from a transmitter into a radio signal
5. signal strength refers to the transmitter power output
6. sorry, not so sure.. Though it might be the waves.