The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 
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First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.
In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.
The electric field E created by the point charge q at any point P, located at a distance r, is defined as:
where K is the constant of Coulomb's law.
In this case, you know:
- K= 9×10⁹
- q= 20 uC=20×10⁻⁶ C
- r= 1 m
Replacing in the definition of electric field:
Solving:
<u><em>E=180000 </em></u><u><em /></u>
Finally, the electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 .
Learn more:
Answer : The main benefits of Nuclear Energy are because it is more efficient than burning most fossil fuels than the amount of energy that is released from the uranium per gram is much more than fuels such as oil and coal. This is approximately eight thousand times more efficient.
Hope this helps you out! ☺
Answer: 
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity 
of the planet P1 with a period 
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding :
(2)
(3)
(4)
On the other hand, we know the gravitational force 
between the star S with mass 
and the planet P1 with mass 
is:
(5)
Where 
is the Gravitational Constant and its value is 
In addition, the centripetal force 
exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding :
(9)
(10)
Finally:
(11) This is the mass of the star S
Galaxy-a collection of dust, gas, AND STARS. But I'm pretty positive this should be the answer.
-- The acceleration of gravity (on Earth) is 9.8 m/s².
-- That means that during every second an object falls,
it adds 9.8 m/s of speed.
Now ! If it adds 9.8 m/s of speed every second, then
how fast is it falling at the end of 3.4 seconds ?
This is as close as I can bring you to the answer
without dropping it at your feet, or handing it to you
on a golden tray.
