A point charge Q produces an electric field of magnitude 80.5 N/C at a distance of 1.08 m. If the field is directed toward the charge, what is the value of q
To find the value of q, we can use Coulomb's Law, which states that the electric field produced by a point charge is given by:
E = k * Q / r^2
Where:
- E is the electric field magnitude,
- k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2),
- Q is the charge producing the field,
- r is the distance from the charge to the point where the field is measured.
In this case, we are given the magnitude of the electric field (E = 80.5 N/C) and the distance (r = 1.08 m). We need to solve for Q.
Rearranging the equation, we have:
Q = E * r^2 / k
Plugging in the given values:
Q = 80.5 N/C * (1.08 m)^2 / (8.99 x 10^9 Nm^2/C^2)
Simplifying the equation, we get:
Q ≈ 1.02986829017 x 10^-6 C
Therefore, the value of q (the charge producing the field) is approximately 1.03 x 10^-6 C.
E=kq/distance
solve for q. q has to be negative, as it is headed toward q.