Determine the magnitude and direction of the electric field that is created at a distance of2.5 mm from a negative point charge of 7.6 nC

To solve this problem, we will use Coulomb's Law, which states that the electric field (E) created by a point charge (q) at a distance (r) is given by:

E = k * |q| / r^2

where k is Coulomb's constant (8.99 * 10^9 N m^2/C^2), |q| is the absolute value of the charge, and r is the distance from the charge. The direction of the electric field is away from the charge if it is positive, and towards the charge if it is negative.

Step 1: Convert the given values to the correct units. The charge q is given as -7.6 nC, which is -7.6 * 10^-9 C. The distance r is given as 2.5 mm, which is 2.5 * 10^-3 m.

Step 2: Substitute the given values into the formula. E = (8.99 * 10^9 N m^2/C^2) * |-7.6 * 10^-9 C| / (2.5 * 10^-3 m)^2

Step 3: Calculate the magnitude of the electric field. E = (8.99 * 10^9 N m^2/C^2) * (7.6 * 10^-9 C) / (2.5 * 10^-3 m)^2 E = 1.08 * 10^6 N/C

Step 4: Determine the direction of the electric field. Since the charge is negative, the direction of the electric field is towards the charge.

So, the magnitude of the electric field created at a distance of 2.5 mm from a negative point charge of 7.6 nC is 1.08 * 10^6 N/C, and the direction is towards the charge.