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:
<h2>
Answer:</h2>
(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
<h2>
Explanation:</h2>
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
A. 4% you divide 2000 to 500 and you get 4 which is your answer
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