To solve the problem, we need to find the distance of the point from the charge and the magnitude of the charge based on the given electric field intensity and electric potential.
1. Given Data:
We are given:
2. Formula for Electric Field and Potential:
For a point charge, the electric field intensity \( E \) and electric potential \( V \) are related to the charge \( q \) and the distance \( r \) as follows:
\[
E = \frac{kq}{r^2}
\]
and
\[
V = \frac{kq}{r}
\]
where \( k \) is Coulomb's constant \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
3. Solving for the Distance \( r \):
From the equation for potential, we can solve for the distance \( r \):
\[
r = \frac{kq}{V}
\]
Substituting the value of \( V = -3600 \, \text{V} \) into the equation gives the distance.
4. Solving for the Charge \( q \):
Now, we can substitute the value of \( r \) into the equation for the electric field \( E = \frac{kq}{r^2} \) to solve for the magnitude of the charge \( q \). This results in the values for the charge and distance of the point from the charge.
Final Answer:
The correct option is (C) \( r = 6 \, \text{m} \), \( q = 2.4 \, \mu\text{C} \).
Two charges, \( q_1 = +3 \, \mu C \) and \( q_2 = -4 \, \mu C \), are placed 20 cm apart. Calculate the force between the charges.