what is the magnitude of the electric field at point 2 m from a point charge q=4nC?
F = k Q q/R^2
E = F/Q = force per unit charge
= k q/R^2
k = 9*10^9 N m^2/C^2
q = 4 * 10^-9 C
R = 2
To find the magnitude of the electric field at a point, you can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric field at a point due to a point charge is given by the equation:
E = k * (|q| / r^2)
where E is the magnitude of the electric field, k is the electrostatic constant (k = 8.99 x 10^9 N·m^2/C^2), |q| is the absolute value of the charge, and r is the distance from the charged object.
In this case, the point charge q is given as 4 nC (nanocoulombs) and the distance r is 2 m. Let's substitute these values into Coulomb's Law:
E = (8.99 x 10^9 N·m^2/C^2) * (|4 nC| / (2 m)^2)
First, convert the charge in nanocoulombs to coulombs:
|q| = 4 nC = 4 x 10^(-9) C
Now, substitute the values into the equation:
E = (8.99 x 10^9 N·m^2/C^2) * (4 x 10^(-9) C / (2 m)^2)
So, the magnitude of the electric field at a point 2 m from a point charge q = 4 nC is given by this final equation. Calculating the values within the equation will give you the answer.
To find the magnitude of the electric field at a point 2 meters from a point charge q=4 nC, you can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric field, E, created by a point charge is given by:
E = (k * q) / r^2
Where:
- E is the electric field magnitude
- k is Coulomb's constant (9 x 10^9 Nm^2/C^2)
- q is the charge of the point charge
- r is the distance from the point charge
Plugging in the values:
E = (9 x 10^9 Nm^2/C^2 * 4 nC) / (2m)^2
Simplifying further:
E = (9 x 10^9 Nm^2/C^2 * 4 x 10^(-9) C) / 4 m^2
E = (36 x 10^9 Nm^2/C) / 4 m^2
E = 9 x 10^9 N/C
Therefore, the magnitude of the electric field at a point 2 meters from a point charge q=4 nC is 9 x 10^9 N/C.