Two point charges +30c and +11c are separated by the distance of 20cm.calculate the electric field intensity and the force on a+5×£-6c charge placed between the charges
To calculate the electric field intensity and the force on a charge placed between two point charges, we can use Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Step 1: Calculate the electric field intensity:
The electric field intensity (E) at a point is defined as the force experienced by a unit positive charge placed at that point. Mathematically, it is given by the formula:
E = F/q
where E is the electric field intensity, F is the force, and q is the test charge.
Step 2: Calculate the force:
The force between two charges is given by the formula:
F = k * (|q1| * |q2|) / r^2
where F is the force, q1 and q2 are the two charges, r is the distance between them, and k is the electrostatic constant (k ≈ 9 x 10^9 Nm^2/C^2).
Let's calculate the values:
Given:
q1 = +30C (charge 1)
q2 = +11C (charge 2)
r = 20cm = 0.2m (distance between charges)
q = +5 × 10^-6C (test charge)
Step 1: Calculate the electric field intensity (E):
E = F/q
Since the test charge is positive, it will experience a force in the same direction as the electric field between the two charges. Therefore, the electric field intensity will be equal to the force between the charges. So, we can find the electric field intensity by calculating the force using Coulomb's law.
Step 2: Calculate the force (F):
F = k * (|q1| * |q2|) / r^2
F = (9 x 10^9 Nm^2/C^2) * ((30C * 11C) / (0.2m)^2)
Now, we can substitute the values and calculate the force.
F = (9 x 10^9 Nm^2/C^2) * (330C^2) / (0.2m)^2
F = (9 x 10^9 Nm^2/C^2) * (330 * C^2) / (0.04 m^2)
F = (9 x 10^9 N) * (330) / (0.04)
F ≈ 7,425,000 N
So, the force experienced by the test charge placed between the two charges is approximately 7,425,000 Newtons.
To calculate the electric field intensity (E), we can now divide the force by the test charge (q).
E = F/q
E = 7,425,000 N / (5 × 10^-6C)
Now, we can substitute the values and calculate the electric field intensity.
E = 7,425,000 N / 5 × 10^-6C
E = 1,485,000,000 N/C
Therefore, the electric field intensity between the two charges is approximately 1,485,000,000 N/C. And the force on the +5 × 10^-6C test charge placed between the charges is approximately 7,425,000 Newtons.