If the total positive charge is Q = 1.62×10−6C , what is the magnitude of the electric field caused by this charge at point P, a distance d = 1.53m from the charge? (Figure 1)
Enter your answer numerically in newtons per coulomb.
To find the magnitude of the electric field caused by the charge, you can use the formula for electric field:
Electric Field (E) = k * (Q / r^2)
where:
- E is the magnitude of the electric field
- k is the electrostatic constant and has a value of 8.99 x 10^9 N m^2/C^2
- Q is the total positive charge
- r is the distance between the charge and the point of interest
In this case, the given values are:
- Q = 1.62 x 10^-6 C
- r = 1.53 m
Substituting the values into the formula, we have:
E = (8.99 x 10^9 N m^2/C^2) * (1.62 x 10^-6 C) / (1.53 m)^2
Calculating this expression will give us the magnitude of the electric field at point P.
To calculate the magnitude of the electric field caused by the positive charge at point P, we can use Coulomb's Law.
Coulomb's Law states that the electric field (E) created by a point charge is given by the equation:
E = k * (Q / r^2)
Where:
- E is the electric field
- k is Coulomb's constant (k ≈ 8.99 x 10^9 N m^2/C^2)
- Q is the total positive charge
- r is the distance between the point charge and the point where the electric field is being measured
Let's substitute the given values into the equation:
Q = 1.62 x 10^-6 C (given)
r = 1.53 m (given)
k ≈ 8.99 x 10^9 N m^2/C^2
E = k * (Q / r^2)
E = (8.99 x 10^9 N m^2/C^2) * (1.62 x 10^-6 C) / (1.53 m)^2
Calculating this expression will give us the magnitude of the electric field at point P.
E = k Q/r^2
direction away from Q since Q is positive