Electric Field
SACE Stage 2 Physics
Electric Fields 1 Solution
1. (a) State Coulomb's Law.
Any two point charges have acting on them equal sized oppositely directed forces acting along the line joining their centres. The magnitude of these equal sized forces is directly proportional to the product of the charges and inversely proportional to the square of their distance apart. The forces are attractive for unlike charges and repulsive for like charges.
[pic].
e.g.
(b) Find the electrostatic force between charges of +2.0 C and +5.0 C separated by a distance of 75 m in a vacuum.
q1 = 2.0 C q2 = 5.0 C r = 75 m
[pic] [pic]
2 Two charges of +8.0 mC and -6.0 mC attract each other with a force of 3.0 x 103 N in a vacuum. What is the distance between the charges?
q1 = 8 x 10-3 C q2 = -6 x 10-3 C
F = 3 x 10 3 N [pic] [pic] [pic]
3 A metal sphere of mass 6.0 x 10-3 kg is found to just float in air above a similar metal sphere when both have a charge of 4.0 mC. Assuming that the only upwards force is electrostatic repulsion, find the distance between the spheres. M = 6.x 10-3 kg g = 9.8 ms-2 q1 = 4 mC
q2 = 4 mC (Could both be + or - charges)
Forces acting have equal magnitudes, opposite directions
Fcoulombic = Fgravitational
Since the gravitaional force is equal and opposite to the coulombic [pic] = mg [pic]
( [pic]
(This indicates how small the gravitational force is compared to the electrostatic!)
4. A charge of 12 (C when placed in an electric field experiences a force of 648 N. What is the magnitude of the electric field strength? q = 12 mC F = 648 N
[pic] [pic] E = 5.4 x 10 -7 N C -1
5. Compare the magnitude of the electrostatic repulsion with the magnitude of the force of gravitational
Electric Fields 1 Solution
1. (a) State Coulomb's Law.
Any two point charges have acting on them equal sized oppositely directed forces acting along the line joining their centres. The magnitude of these equal sized forces is directly proportional to the product of the charges and inversely proportional to the square of their distance apart. The forces are attractive for unlike charges and repulsive for like charges.
[pic].
e.g.
(b) Find the electrostatic force between charges of +2.0 C and +5.0 C separated by a distance of 75 m in a vacuum.
q1 = 2.0 C q2 = 5.0 C r = 75 m
[pic] [pic]
2 Two charges of +8.0 mC and -6.0 mC attract each other with a force of 3.0 x 103 N in a vacuum. What is the distance between the charges?
q1 = 8 x 10-3 C q2 = -6 x 10-3 C
F = 3 x 10 3 N [pic] [pic] [pic]
3 A metal sphere of mass 6.0 x 10-3 kg is found to just float in air above a similar metal sphere when both have a charge of 4.0 mC. Assuming that the only upwards force is electrostatic repulsion, find the distance between the spheres. M = 6.x 10-3 kg g = 9.8 ms-2 q1 = 4 mC
q2 = 4 mC (Could both be + or - charges)
Forces acting have equal magnitudes, opposite directions
Fcoulombic = Fgravitational
Since the gravitaional force is equal and opposite to the coulombic [pic] = mg [pic]
( [pic]
(This indicates how small the gravitational force is compared to the electrostatic!)
4. A charge of 12 (C when placed in an electric field experiences a force of 648 N. What is the magnitude of the electric field strength? q = 12 mC F = 648 N
[pic] [pic] E = 5.4 x 10 -7 N C -1
5. Compare the magnitude of the electrostatic repulsion with the magnitude of the force of gravitational