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Calculate the force between two point charges of 20 and 40 Coulombs that are separated by 0.8 meters.

Chapter 32 final exam review Charge originates from protons and electrons in atoms. Electric force—caused by interactions between charges. F = k q 1 q 2 d 2 where k = 9 x 10 9 N m 2 /C 2 . q e = 1.6 x 10 -19 C.

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Calculate the force between two point charges of 20 and 40 Coulombs that are separated by 0.8 meters.

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  1. Chapter 32 final exam reviewCharge originates from protons and electrons in atoms.Electric force—caused by interactions between charges.F = k q1 q2 d2where k = 9 x 109 N m2/C2.qe = 1.6 x 10-19 C.

  2. Calculate the force between two point charges of 20 and 40 Coulombs that are separated by 0.8 meters. A point charge of 10 Coulombs experiences a force of 0.14 N when another charge is placed 0.02 meters from it. Calculate the magnitude of the other charge.

  3. Calculate the distance between two point charges of 5 and 8 Coulombs that experience a force of 0.012 N. Two identical charges experience a force of 2.5 N when they are placed 0.45 m apart. Calculate the magnitude of the two charges.

  4. Three charges are placed as shown below. Calculate the force experienced by each of the charges. 6 C -8 C -3 C .8 m 1.2 m

  5. Chapter 33 final exam review How are electric fields and forces related? Field lines flow out of positive charges. Gravitational potential energy = mgh Electric potential energy = qEd Work is equal to energy. Electric potential is the potential energy per unit charge = qEd/q = Ed = Voltage.

  6. A charge of 5.3 C is placed in an electric field with a strength of 27.9 N/C. Calculate the force experienced by the charge. A charge of 2.7 C is placed in an electric field that has a strength of 27.4 N/C. Calculate the force that the charge experiences.

  7. An electric field of 45.8 N/C is directed upward from the bottom of this page. If a charge of -25 C is placed in this uniform field, calculate the magnitude and direction of the electrical force experienced by the charge.

  8. Calculate the amount of work required to move an electron 3.5 m against a uniform electrical field of 250 N/C. Two charged plates are separated by a distance of 0.014 m. If a potential difference of 24 V is applied to the plates, calculate the strength of the electric field between the plates.

  9. Chapter 34 final exam review Current is the flow of charge as a result of potential difference. Current I = q / t. Ohm’s law V = I R Power = current x voltage. P = VI = I2R = V2/R Pg 546 P/C #1-8, T/S # 1-5

  10. Chapter 36/37 final exam review Magnetic fields create lines of force that wrap around the poles of the magnet. Current carrying wires produce magnetic fields. The direction can be determined using the right hand rule. Magnets deflect charges and current carrying wires.

  11. Chapter 36/37 final exam review Electromagnetic induction Generator—coil of wire moves in a magnetic field. Transformers—current carrying wires wrapped around a magnet to increase or decrease voltage.

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