What is the value and the direction of an electric field at a distance of 2.5m from a +1 nC charge?
A) approximately 1.45 V/m directed toward the positive charge
B) approximately 1.45 V/m directed away from the positive charge
C) approximately 145 V/m directed away from the positive charge
D) approximately 145 V/m directed toward the positive charge
The correct answer is D) approximately 145 V/m directed toward the positive charge.
The electric field around a point charge is given by the equation:
E = (kQ) / r^2
Where:
E is the electric field
k is the Coulomb constant (k = 8.99 x 10^9 N m^2/C^2)
Q is the charge of the point charge
r is the distance from the point charge to the location where the electric field is being measured.
In this case, the charge (Q) is +1 nC = +1 x 10^-9 C, and the distance (r) is 2.5 m. Plugging these values into the equation:
E = (8.99 x 10^9 N m^2/C^2 * 1 x 10^-9 C) / (2.5 m)^2
E = 8.99 x 10^9 N m^2/C^2 / 6.25 m^2
E ≈ 1.44 x 10^9 N/C
Since the charge is positive, the electric field points away from the charge. Therefore, the direction of the electric field is towards the positive charge.
Therefore, the correct answer is D) approximately 145 V/m directed toward the positive charge.