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 .
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Yes I can help jus tell me what the 26th question is
During the first phase of acceleration we have:
v o = 4 m/s; t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?; v o = 13 m/s; a = 1.125 m/s² ; t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s
