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Physics Problem Solution: Force and Field Equations

Solve physics problems involving forces between charges, free-body diagrams, and field strength calculations. Learn how to calculate resultant forces and electric field strength. Detailed solutions provided.

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Physics Problem Solution: Force and Field Equations

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  1. Write down an equation for the force between two point charges, Q1 and Q2 , separated by a distance r (1) A speck of dust has a mass of 1.0 × 10–18 kg and carries a charge equal to that of one electron. Near to the Earth’s surface it experiences a uniform downward electric field of strength100 N C–1 and a uniform gravitational field of strength 9.8 N kg–1. Draw a free-body force diagram for the speck of dust. Label the forces clearly. Calculate the magnitude and direction of the resultant force on the speck of dust. (6) (Total 7 marks)

  2. A speck of dust has a mass of 1.0 × 10–18 kg and carries a charge equal but opposite to that of one electron. Near to the Earth’s surface it experiences a uniform downward electric field of strength100 N C–1 and a uniform gravitational field of strength 9.8 N kg–1. Draw a free-body force diagram for the speck of dust. Label the forces clearly mass of 1.0 × 10–18 kg Charge 1.6 x 10-19C Fe Near to the Earth’s surface it experiences a uniform downward electric field of strength Fg Where Fgis the force due to the gravitational field and Fe is the force due to the electric field

  3. Calculate the magnitude and direction of the resultant force on the speck of dust. mass of 1.0 × 10–18 kg Charge 1.6 x 10-19C Fe Fg Resultant force Fg+ Fe

  4. Calculate the magnitude and direction of the resultant force on the speck of dust. mass of 1.0 × 10–18 kg Charge 1.6 x 10-19C The force due to the field (Fe) The force due to gravity (Fg) F=mg (where g = 9.8Nkg-1) Fg = 1.0 x 10-18 x 9.8 Fg= 9.8 x 10-18 N From Resultant force Fg+ Fe The resultant force Fg + Fe= (98 +160) x 10-19 = 258 x 10-19N

  5. The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. The oil drop has a mass 9.79 x 10–15 kg. The potential difference between the plates is 5000 V and plate B is at a potential of 0 V. Is plate A positive or negative? ……………………………………………………………………………………………… Draw a labelled free-body force diagram which shows the forces acting on the oil drop.(You may ignore upthrust). (3) Calculate the electric field strength between the plates. Electric field strength =………………………………… (2) Calculate the magnitude of the charge Q on the oil drop. Charge =…………………………………… How many electrons would have to be removed from a neutral oil drop for it to acquire this charge? ……………………………………………………………………………………………… (3) (Total 8 marks)

  6. The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. Here because the body is at rest the resultant force must be zero. Fe = Fg Fe Free body diagram Fg

  7. The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. Is plate A positive or negative?

  8. The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. __ __ __ Fe Plate A has to have a negative charge as the drop would accelerate down if the field were reversed + Fg

  9. The potential difference between the plates is 5000 V and plate B is at a potential of 0 V. Calculate the electric field strength between the plates. The equation for the field strength in a UNIFORM FIELD is The potential difference between the plates V= 5000V. The distance between the plates is 2.50 x 10-2m

  10. Calculate the magnitude of the charge Q on the oil drop. Fe We want to use Fg We now know that the field has this value The force due to the field is equal to the force of gravity on the oil drop as it is at rest: F=mg (where m = 9.79 x 10–15 kg and g =9.81Nkg-1) F= 9.79 x 10–15 x 9.81 = 9.60 x10-14 N This value is 3 electrons worth of charge

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