Question

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) ( 2.25 X 10⁵ C ); (b) ( 9.00 N/C )
b) ( 2.25 X 10⁴ C ); (b) ( 9.00 N/C )
c) ( 2.25 X 10⁵ C ); (b) ( 900.00 N/C )
d) ( 2.25 X 10⁴ C ); (b) ( 900.00 N/C )

Answer 1

To create a 10,000 N/C electric field at a distance of 0.250 m, a magnitude point charge of 2.25 x 10^(-4) C is required. At a distance of 10.0 m, the electric field is 9.00 N/C.

Step-by-step explanation:

To find the magnitude of the point charge that creates a 10,000 N/C electric field at a distance of 0.250 m, we can use the formula for electric field:

Electric field (E) = k * (magnitude of point charge (q)) / (distance squared)

Plugging in the given values, we have:

10,000 N/C = (9 * 10^9 N m^2/C^2) * (q) / (0.250 m)^2

Solving for q, the magnitude of the point charge, we get:

q = (10,000 N/C) * (0.250 m)^2 / (9 * 10^9 N m^2/C^2) = 2.25 * 10^(-4) C

Therefore, the correct answer is:

(a) (2.25 x 10^(-4) C)

To find the electric field at a distance of 10.0 m, we can use the same formula for electric field:

Electric field (E) = k * (magnitude of point charge (q)) / (distance squared)

Plugging in the given values, we have:

Electric field (E) = (9 * 10^9 N m^2/C^2) * (2.25 x 10^(-4) C) / (10.0 m)^2 = 9.00 N/C

Therefore, the correct answer is:

(b) (9.00 N/C)