Book page. Coulombs Law

Transcription

1 Book page Coulombs Law

2 A Coulomb torsion balance A Coulomb torsion balance is used to measure the force between two charged objects Coulomb's Torsion Balance Two conducting spheres fixed on insulating rod, suspended by a thin wire Enclosed in container to avoid air movement Test charge is lowered Conclusion: F q 1 q 2 and F between point charges 1 R 2 F = kq 1q 2 = q 1q 2 where k = 9.9 R 2 4πε 0 R and ε 0 =

3 No need for sign convention In force calculations there is no need to use sign convention Use magnitude of point charge Draw diagram with sign of each point charge This will indicate whether force is attractive or repulsive One Coulomb of charge is the charge on electrons or protons Q e/p = cgrahamphysics.com C 2016

4 Electric and gravitational force Similarities between F g and F e Gravitational force Between two pint masses Electric Force Between two point charges (in vacuum) Product is the inverse Law F g = GMm R 2 F e = kq 1q 2 R 2 Difference between F g and F e No gravitational analogue to electrical permittivity ε 0 F g is always attractive

5 Electric field (Faraday field) A region in space under the influence around a point charge or a group of point charges The stronger the field, the stronger the force that is experienced The direction of the electric field is defined as the direction of a force it causes to act on a small positive test charge Lines of electric flux or lines of force in an electric field are obtained by the direction an isolated positive test charge would follow if placed at a point in the field + +

6 Rules for field lines Field lines are imaginary lines to help us understanding the nature of electric fields 1. Start on positive, end on negative 2. Meet electrostatically charged conductors at right angles 3. They never cross 4. Their density indicates the strength of the electric field 5. No electric field exists on a hallow conductor 6. Electric field is uniform between tow oppositely charged conducting plates

7 Electric field strength The electric field strength or electric field intensity at any point in space, E, is equal to the force per unit charge exerted on a positive test charge E = F or F = QE Q E is a vector in N C 1 or Vm 1 The electric field due to a point charge is defined as E = F kqq = R 2 Q Q = kq R 2

8 Example Find the force acting between two point charges of +10μC and - 5 μc separated by a distance of 10 cm in vacuum Solution F e = kq 1q 2 R 2 = = 45N Since the force is negative, there is a force of attraction between the point charges We could just have used the magnitude and realized from the sign of the charges that they attract

9 Example The force between two point charges is 20.00N. If one charge is doubled, the other charge tripled and the distance between them is halved, calculate the resultant force between them Solution Write down everything given: Q 1 = 2Q 1, Q 2 = 3Q 2, R = 1 R and F = 20N 2 Use ratios F = F kq1q2 R 2 = k2q13q2 1 4 R = 1 24 F = 1 24 F or F = 24F = = 480N

10 Example Charges of +1C are located at the corners of a 45 degree right angled triangle. Determine the resultant force on the charge located at the right angle Solution Draw the force experienced at the required position F 2 = kq 1q 2 = = N R F 3 = kq 1q 3 = = N R Find the resultant force using Pythagorean Theorem: F R = = N Find required direction using trigonometry: tan θ = cgrahamphysics.com = θ =

11 Example Consider the same problem as before, but the charges are set up differently. Determine the resultant force on the charge located at the right angle Solution F 2 = kq 1q 2 R 2 = = N F 3 = N However, the resultant force is different: The resultant force is N vertically down

12 Example A point charge of 25 μc experiences a force of N. Calculate the electric field strength producing this force. Solution E = F = = 4.0NC 1 in the direction Q of the force

13 Example Calculate the electric field strength 1.5 cm from a point charge of 100 pc in a vacuum. Solution E = q = 4πε 0 R π = This is about NC 1 radially outward

14 Example The field at a particular place due to more than one point charge is the vector sum of the fields caused by each point charge on its own. Calculate the electric field on X due to the charges shown. Solution E 2 = kq 2 R 2 = = E 1 = kq 1 R 2 = = R = = NC 1 θ = tan 1 E 2 E 1 θ E 1 R E 2