What is the magnitude and direction of the electric field at a point 0.75 m away from a point charge of +2.0 pC?
Have you considered coulombs law?
E=kq/r^2
happy new yr
To find the magnitude and direction of the electric field at a point, we can use Coulomb's Law. Coulomb's Law states that the electric field created by a point charge is directly proportional to the charge and inversely proportional to the distance from the point charge.
The formula for the electric field due to a point charge is:
E = k * q / r^2
Where:
E is the electric field
k is the electrostatic constant (9.0 x 10^9 N⋅m^2/C^2)
q is the charge of the point charge
r is the distance from the point charge
In this case, the charge of the point charge is given as +2.0 pC (picocoulombs) which can be converted to C (coulombs) by dividing by 10^12, which means q = +2.0 x 10^-12 C. And the distance from the point charge is given as 0.75 m, which means r = 0.75 m.
Now we can substitute the values into the formula to calculate the magnitude of the electric field.
E = (9.0 x 10^9 N⋅m^2/C^2) * (+2.0 x 10^-12 C) / (0.75 m)^2
Simplifying the equation gives:
E = 2.4 x 10^5 N/C
Therefore, the magnitude of the electric field at a point 0.75 m away from the +2.0 pC point charge is 2.4 x 10^5 N/C.
To determine the direction of the electric field, we can use the principle that positive charges repel each other. Since the point charge is positive, the electric field will point away from the charge in all directions.