What magnitude point charge creates a 10,000 N/C electric field at a distance of 0.250 m? (b) How large is the field at 10.0 m?

(a) 69.52 x 10⁻⁹ C

(b) 6.25 N/C

Explanation:

(a) The electric field (E) due to a point charge is directly proportional to the magnitude of the point charge (Q) and inversely proportional to the square of the distance (r) between the point charge and the point where the electric field is. This can be represented as follows;

E = k Q / r²         (i)

Where;

k = constant of proportionality called electric constant = 8.99 x 10⁹Nm²/C²

From the question, the following are given;

E = 10000N/C

r = 0.250m

Substitute these values into equation (i) as follows;

=> 10000 = 8.99 x 10⁹ x Q / 0.250²

=> 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 = 69.52 x 10⁻⁹ C

Therefore, the magnitude of the point charge is 69.52 x 10⁻⁹ C.

(b) By the same token, to calculate the magnitude of the electric field at 10.0m, substitute the values of Q = 69.52 x 10⁻⁹ C, k = 8.99 x 10⁹ and r = 10.0m into equation (i) as follows;

=> E = 8.99 x 10⁹ x 69.52 x 10⁻⁹ / 10.0²

=> E = 8.99 x 69.52 / 100.0

=> E = 6.25 N/C

Therefore, the electric field at 10.0m is as large as 6.25 N/C

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according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

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