Two point charges of magnitude 2.0 X 10^-7 C and 85 X 10^-8 C are placed 12 cm apart .What is the field intensity of each charge at the site of other ?
To find the electric field intensity at the site of one charge due to the other charge, we can use Coulomb's Law. Coulomb's Law states that the electric field intensity (E) between two point charges is given by:
E = k * (|q1| / r^2),
where
E is the electric field intensity,
k is the electrostatic constant (9.0 x 10^9 Nm^2/C^2),
q1 is the magnitude of the first charge,
r is the distance between the charges.
Let's denote the first charge with magnitude 2.0 x 10^-7 C as q1, and the second charge with magnitude 85 x 10^-8 C as q2.
To find the electric field at the site of q1 due to q2, we consider q2 as the source charge and q1 as the test charge. So, the electric field intensity at the location of q1 due to q2 can be calculated using Coulomb's Law.
First, we need to calculate the distance between the charges. The problem states that the charges are placed 12 cm apart, which is equivalent to 0.12 m.
Now, let's calculate the electric field intensity at the site of q1 due to q2:
E2 = k * (|q2| / r^2)
= (9.0 x 10^9 Nm^2/C^2) * (85 x 10^-8 C) / (0.12 m)^2
E2 = 2.395 x 10^4 N/C
Therefore, the field intensity of the second charge at the site of the first charge is 2.395 x 10^4 N/C.