Two insulated charged metallic spheres A & B have their centres separated by a distance of 50cm in air. Calculate the force of electrostatic repulsion between them. If the charge on each is +5.0*10^-8c. The radii of A & B are negligible compared to their distance of separation.
( b). What is the force of repulsion if (i)the charge doubled and between them is halved.
(ii) the two spheres are placed in water of dielectric constant 81? [(a)F=9*10^-5N,(b)(i)F=1.4*10^-3N, (ii)F=1.1*10^-6N]
a) use coulomb's law and your calculator.
b) Double charges? force goes up by 2*2. Distance is halved? Force goes up by 4
total force increase: 4*4
c) dielectric constant will reduce the force by a factor of 1/81
I didn't check your answers.
To calculate the force of electrostatic repulsion between the two spheres, we can use Coulomb's Law, which states that the force between two charged objects is given by:
F = k * (|q1| * |q2|) / (r^2)
where F is the force, k is the electrostatic constant (9 * 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges on the spheres, and r is the distance between their centers.
Given:
Distance between the centers, r = 50 cm = 0.5 m.
Charge on each sphere, |q1| = |q2| = 5.0 * 10^-8 C.
k = 9 * 10^9 Nm^2/C^2.
Substituting the values into the formula:
F = (9 * 10^9 Nm^2/C^2) * (5.0 * 10^-8 C * 5.0 * 10^-8 C) / (0.5 m)^2
Calculating the above expression:
F = 9 * 10^9 Nm^2/C^2 * 25 * 10^-16 C^2 / 0.25 m^2
= 9 * 25 * 10^-7 N
= 9 * 10^-6 N
= 9 * 10^-5 N (approx.)
Therefore, the force of electrostatic repulsion between the two spheres is approximately 9 * 10^-5 N.
(b) (i) If the charge is doubled (i.e., 2 * 5.0 * 10^-8 C) and the distance is halved (i.e., 0.25 m), we can apply Coulomb's Law again to calculate the new force.
F' = (9 * 10^9 Nm^2/C^2) * (2 * 5.0 * 10^-8 C * 2 * 5.0 * 10^-8 C) / (0.25 m)^2
Calculating the above expression:
F' = 9 * 10^9 Nm^2/C^2 * 100 * 10^-16 C^2 / 0.0625 m^2
= 9 * 100 * 10^-7 N
= 9 * 10^-5 N
= 1.4 * 10^-3 N (approx.)
Therefore, if the charge is doubled and the distance is halved, the new force of repulsion is approximately 1.4 * 10^-3 N.
(ii) If the spheres are placed in water of dielectric constant 81, Coulomb's Law needs to be modified using the concept of relative permittivity.
F'' = (k * (|q1| * |q2|) / (εr * r^2)
where εr is the relative permittivity (dielectric constant) of the medium.
Given:
εr = 81.
Substituting the values into the formula:
F'' = (9 * 10^9 Nm^2/C^2) * (5.0 * 10^-8 C * 5.0 * 10^-8 C) / (81 * 0.5 m^2)
Calculating the above expression:
F'' = 9 * 10^9 Nm^2/C^2 * 25 * 10^-16 C^2 / (81 * 0.25 m^2)
= 9 * 25 * 10^-7 N / (81/4)
= 9 * 25 * 10^-7 * 4/81 N
= 9 * 100 * 10^-7 / 81 N
= 9 * 1000 * 10^-9 / 81 N
= 9 * 10^-6 * 10^-9 / 9
= 10^-15 N / 9
= 1.111... * 10^-16 N (approx.)
Therefore, if the spheres are placed in water with a dielectric constant of 81, the force of repulsion is approximately 1.1 * 10^-16 N.