Question

A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength equal to 5.8 × 105 newtons/coulomb?

Answer 1

The distance from the ball's center where the electric field strength is equal to 5.8 × 10^5 N/C is approximately 3.03 meters.

Step-by-step explanation:

The strength of an electric field (E) can be calculated using the equation E = kQ/r^2, where k is the electrostatic constant (k = 8.99 × 10^9 N・m^2/C^2), Q is the charge of the object, and r is the distance from the object's center. In this case, the charge of the ball is +1.5 × 10^-9 C and we need to find the distance at which the electric field strength is 5.8 × 10^5 N/C.

Using the equation, we can rearrange it to solve for r: r = sqrt(kQ/E).

Plugging in the values, we get: r = sqrt((8.99 × 10^9 N・m^2/C^2)(1.5 × 10^-9 C)/(5.8 × 10^5 N/C)) = 3.03 meters.

Therefore, the distance from the ball's center where the electric field strength is equal to 5.8 × 10^5 N/C is approximately 3.03 meters.

Answer 2

F = kq1q2/r2
Where,
F - Coulomb Force
k - constant value which is equal to 8.98 × 10^9 newton square metre per square coulomb
q1 and q2 - two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
5.8 * 10^5 r^2 = 1.5*10^-9
r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3