What is the magnitude and direction of the electric field due to a -5.8 μC point charge at a distance of 9.8 m?*1 point5.326 kN/C towards the charge0.544 kN/C towards the charge5.686 kN/C away from the charge0.533 kN/C away from the charge
Question
What is the magnitude and direction of the electric field due to a -5.8 μC point charge at a distance of 9.8 m?*1 point5.326 kN/C towards the charge0.544 kN/C towards the charge5.686 kN/C away from the charge0.533 kN/C away from the charge
Solution
The magnitude of the electric field (E) due to a point charge (q) at a distance (r) can be calculated using Coulomb's Law, which states:
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.
Step 1: Convert the charge from μC to C. 1 μC = 10^-6 C, so -5.8 μC = -5.8 * 10^-6 C.
Step 2: Substitute the values into the equation:
E = (8.99 * 10^9 N m^2/C^2) * |-5.8 * 10^-6 C| / (9.8 m)^2 E = (8.99 * 10^9 N m^2/C^2) * 5.8 * 10^-6 C / 96.04 m^2 E = 0.544 kN/C
The direction of the electric field is always from positive to negative charges. Since the charge in this case is negative, the electric field is towards the charge.
So, the magnitude and direction of the electric field due to a -5.8 μC point charge at a distance of 9.8 m is 0.544 kN/C towards the charge.
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