# A glass rod is charged positive by rubbing it with (artificial) fur

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A glass rod is charged positive by rubbing it with (artificial) fur
A glass rod is charged positive by rubbing it with (artificial) fur. It is then used to charge the electroscope as follows: the rod is brought near to the electroscope terminal, causing a deflection. While the rod is held in place, a grounded wire is touched to its terminal, discharging the electroscope. The glass rod is then withdrawn, resulting in a deflection of the electroscope. That same glass rod is then charged positive in the same way and brought near to the terminal of the electroscope. How will the electroscope respond? When the charged glass rod is brought near the terminal, the electroscope will: (a) indicate a higher voltage. (b) indicate a lower voltage. (b) indicate a lower voltage. (c) remain the same.

How do charges interact?
If you have two charges of any sign placed close to each other and isolated from the rest of the universe, what could happen? Charges could push each other away – Repulsive Force One charges could pull the other towards itself – Attractive Force These are not contact forces!

Attractive and Repulsive Forces
Are each of the following attractive or repulsive forces? Repulsive Repulsive Attractive Attractive

1. 1 and 3 carry charges of opposite sign.
Three pithballs are suspended from thin threads. Various objects are then rubbed against other objects (nylon against silk, glass against polyester, etc.) and each of the pithballs is charged by touching them with one of these objects. It is found that pithballs 1 and 2 repel each other and that pithballs 2 and 3 repel each other. From this we can conclude that 1. 1 and 3 carry charges of opposite sign. 2. 1 and 3 carry charges of equal sign. 3. all three carry the charges of the same sign. 4. one of the objects carries no charge. 5. we need to do more experiments to determine the sign of the charges. Answer: 3. Charges of equal sign repel, so 1 and 2 carry charges of equal sign and so, too, do 2 and 3.

1. 1 and 3 carry charges of opposite sign.
Three pithballs are suspended from thin threads. Various objects are then rubbed against other objects (nylon against silk, glass against polyester, etc.) and each of the pithballs is charged by touching them with one of these objects. It is found that pithballs 1 and 2 attract each other and that pithballs 2 and 3 repel each other. From this we can conclude that 1. 1 and 3 carry charges of opposite sign. 2. 1 and 3 carry charges of equal sign. 3. all three carry the charges of the same sign. 4. one of the objects carries no charge. 5. we need to do more experiments to determine the sign of the charges. Answer: 5. Charges of opposite sign attract, charges of equal sign repel, and any type of charge attracts a neutral object. So 1 and 2 either carry charges of opposite sign or one of the two is neutral and the other charged. Since 2 and 3 repel, however, we know that 2 and 3 carry charges of equal sign. So there are two possibilities: 2 and 3 carry charges of equal sign and (i) 1 is neutral, or (ii) 1 carries a charge of the opposite sign to that of 2 and 3.

Determination of the magnitude of attractive or repulsive forces between point charges
Coulomb’s Law Point charge – charged object with negligible dimensions = magnitude of first point charge = magnitude of second point charge r12 = separation distance between q1 and q2 ke = Proportionality constant = 8.99 x 109 = Magnitude of electric force that q1 exerts on q2 e0 = Permittivity of free space = 8.85 x 10-12 Permittivity - How easily charge flows through free space Different materials have their own permittivity

Determination of direction of attractive or repulsive forces
q2 q1 F21 = -F12 F21 F12 F21 F12 q1 q2 F21 = -F12

1. Yes, we must move the particles farther apart.
A hydrogen atom is composed of a nucleus containing a single proton, about which a single electron orbits. The electric force between the two particles is 2.3 x 1039 greater than the gravitational force! If we can adjust the distance between the two particles, can we find a separation at which the electric and gravitational forces are equal? 1. Yes, we must move the particles farther apart. 2. Yes, we must move the particles closer together. 3. No, at any distance Answer: 3. Both the electric and gravitational forces vary as the inverse square of the separation between two bodies. Thus, the forces cannot be equal at any distance.

Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces: Answer: 5. The magnitude of the electrostatic force exerted by 2 on 1 is equal to the magnitude of the electrostatic force exerted by 1 on 2. If the charges are of the same sign, the forces are repulsive; if the charges are of opposite sign, the forces are attractive.

Example: Two charges, one positive and one negative are separated by a distance of 2 cm. The positive charge has a magnitude of 2 C and the negative charge has a magnitude of 2 C What is the magnitude and direction of the force on charge 2 by charge 1? What is the magnitude and direction of the force on charge 1 by charge 2? What happens to the magnitude and direction of the answer from a) if the positive charge is doubled? What happens to the magnitude and direction of the answer from b) if the negative charge is doubled? 2 cm q2 q1

This is an example of Newton’s 3rd Law!!!
2 cm F21 F12 q2 q1 = 8.99 x 1013 N towards q1 = 8.99 x 1013 N towards q2 If q1 is doubled to 4 C the magnitude of the force is also doubled to 1.80 x 1014 N, but the direction doesn’t change If q2 is doubled to 4 C the magnitude of the force is also doubled to 1.80 x 1014 N, but the direction doesn’t change This is an example of Newton’s 3rd Law!!!

Interactions between more than two charges
Forces are vectors and must be combined using vector addition. You will typically examine the net force on one of the charges present. F12 F32 q3 q1 q2 r12 r23 Vector form of Coulomb’s Law F2 = F12 + F32

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