Coulomb Law Practice Problems Solutions

1. Determine the quantity of charge on … 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑐ℎ𝑎𝑟𝑔𝑒=𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑒𝑟 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 × 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 a. … a plastic tube which has been rubbed with animal fur and gained 3.8x109 electrons. 𝐶ℎ𝑎𝑟𝑔𝑒= 1.60× 10 −19 3.8× 10 9 =6.1 x 10−10 C (of negative charge) b. … a vinyl balloon which has been rubbed with animal fur and gained 1.7x1012 electrons. 𝐶ℎ𝑎𝑟𝑔𝑒= 1.60× 10 −19 1.7× 10 12 =2.7 x 10−7 C (of negative charge) c. … an acetate strip which has been rubbed with wool and lost 7.3x108 electrons. 𝐶ℎ𝑎𝑟𝑔𝑒= 1.60× 10 −19 7.3× 10 8 =1.2 x 10−10 C (of positive charge)

2. Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on the balls are -3.1x10-7 C and -3.7x10-7 C. Determine the force of electrical repulsion when held a distance of 42 cm apart. 𝐹=𝑘 𝑞 1 𝑞 2 𝑑 2 =9.0× 10 9 (−3.1× 10 −7 )(−3.7× 10 −7 ) (0.42) 2 =5.8× 10 −3 𝑁

3. Mr. H gives two large vinyl balloons ten good rubs on what's left of his hair, transferring a total of 2.1x1012 electrons from his hair to each balloon. He walks away, leaving the balloons to be held by strings from a single pivot point on the ceiling. The balloons repel and reach an equilibrium position with a separation distance of 58 cm. a. Determine the quantity of charge on each balloon. 𝐶ℎ𝑎𝑟𝑔𝑒= 1.60× 10 −19 2.1× 10 12 =3.4× 10 −7 C (of negative charge) b. Determine the Coulomb force of repulsion between the two balloons. 𝐹=𝑘 𝑞 1 𝑞 2 𝑑 2 =9.0× 10 9 (−3.4× 10 −7 )(−3.4× 10 −7 ) (0.58) 2 =3.0× 10 −3 𝑁

4. Two different objects are given charges of +3.27 μC and -4.91 μC. What separation distance will cause the force of attraction between the two objects to be 0.358 N? (GIVEN: 1 C = 106 μC) 𝑑 2 =𝑘 𝑞 1 𝑞 2 𝐹 =9.0× 10 9 (3.27× 10 −6 )(−4.91× 10 −6 ) −0.358 =0.403 𝑑= 0.403 =0.635 𝑚

5. Two objects with charges Q1 and Q2 experience an electrical force of attraction of 8.0x10-4 N when separated by a distance of d. Determine the force of attraction if the same objects are separated by … a. … a distance of 2•d.  ¼ of the force  ¼ (8.0 x 10-4) = 2.0 x 10-4 N b. … a distance of 3•d.  1/9 of the force  1/9 (8.0 x 10-4) = 8.9 x 10-5 N c. … a distance of 0.5•d.  4 x the force  4 (8.0 x 10-4) = 3.2 x 10-3 N d. … a distance of 2d and each object having double the charge.  remain the same 8.0 x 10-4 N

6. The electric field intensity at a particular location surrounding a Van de Graaff generator is 4.5x103 N/C. Determine the magnitude of the force which this field would exert upon … a. … an electron when positioned at this location. 𝐸= 𝐹 𝑞  𝐹=𝐸𝑞= 4.5× 10 3 1.6× 10 −19 =7.2× 10 −16 𝑁 b. … a charged balloon with 1.8 μC of charge when positioned at this location. 𝐸= 𝐹 𝑞  𝐹=𝐸𝑞= 4.5× 10 3 1.8× 10 −6 = 8.1x10-3 N c. … a pith ball with 6.8x10-8 C of charge when positioned at this location. 𝐸= 𝐹 𝑞  𝐹=𝐸𝑞= 4.5× 10 3 6.8× 10 −8 =3.1× 10 −4 N

7. A test charge with a negative charge of 2.18x10-8 C experiences a northward force of 4.50x10-5 N when placed a distance of 25.0 cm from a source charge. a. Determine the magnitude and direction of the electric field at this location. 𝐸= 𝐹 𝑞 = 4.50× 10 −5 2.18× 10 −8 =2.06× 10 3 N/C b. Determine the magnitude and type of charge on the source. 𝐸= 𝑘𝑄 𝑑 2  𝑄= 𝐸 𝑑 2 𝑘 = 2.06× 10 3 (.25) 2 9× 10 9 =1.43× 10 −8 𝐶 c. Determine the strength of the electric field at a distance of 75.0 cm from the source. 𝐸= 𝑘𝑄 𝑑 2 = (9× 10 9 )(1.43× 10 −8 ) (0.75) 2 =2.28× 10 2 𝑁/𝐶

8. The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determine the force and resulting acceleration of an electron (m = 9.11x10-31 kg) as it travels through this electric field towards the television screen. 𝐸= 𝐹 𝑞  𝐹=𝐸𝑞= 2.5× 10 4 1.6× 10 −19 =4× 10 −15 𝑁 𝐹=𝑚𝑎  𝑎= 𝐹 𝑚 = 4× 10 −15 9.11× 10 −31 =4.39 𝑚 𝑠 2

9. Consider the diagram below. GIVEN: Q1 = +5.00x10-7 C; Q2 = +4.00x10-7 C; Q3 = -8.00x10-7 C; d1 = 5.00 cm; d2 = 8.00 cm. a. Determine the magnitude and direction of the force exerted by Q2 upon Q1. 𝐹 2 𝑜𝑛 1 =𝑘 𝑞 1 𝑞 2 𝑑 2 =9.0× 10 9 (5.00× 10 −7 )(4.00× 10 −7 ) (0.05) 2 =0.72 𝑁 𝑙𝑒𝑓𝑡𝑤𝑎𝑟𝑑 (𝑟𝑒𝑝𝑢𝑙𝑠𝑖𝑜𝑛) b. Determine the magnitude and direction of the force exerted by Q3 upon Q1. 𝐹 3 𝑜𝑛 1 =𝑘 𝑞 1 𝑞 2 𝑑 2 =9.0× 10 9 (5.00× 10 −7 )(−8.00× 10 −7 ) (0.05+0.08) 2 =−0.21 𝑁 𝑟𝑖𝑔ℎ𝑡𝑤𝑎𝑟𝑑 (𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛) c. Determine the magnitude and direction of the net electric force on Q1. 𝐹 𝑛𝑒𝑡 = 𝐹 2 𝑜𝑛 1 + 𝐹 3 𝑜𝑛 1 =0.72+ −.21 =0.51 𝑁

10. Three charges are arranged as shown in the diagram at the right. GIVEN: Q1 = -15 nC Q2 = +14 nC Q3 = +11 nC 1 nC = 1 nano Coulomb = 1x10-9 C a. Determine the magnitude and direction of the force exerted by Q1 upon Q2. 𝐹 2 𝑜𝑛 1 =𝑘 𝑞 1 𝑞 2 𝑑 2 =9.0× 10 9 (−15× 10 −9 )(14.00× 10 −9 ) (0.70) 2 =−3.9× 10 −6 𝑁 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 (𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛) b. Determine the magnitude and direction of the force exerted by Q3 upon Q2. 𝐹 3 𝑜𝑛 2 =𝑘 𝑞 2 𝑞 3 𝑑 2 =9.0× 10 9 (14× 10 −9 )(11.00× 10 −9 ) (0.30) 2 =1.5× 10 −5 𝑁 𝑙𝑒𝑓𝑡𝑤𝑎𝑟𝑑 (𝑟𝑒𝑝𝑢𝑙𝑠𝑖𝑜𝑛)

10. c. Determine the magnitude and direction of the net electrostatic force on Q2. 𝐹 𝑛𝑒𝑡 = (1.5× 10 −5 ) 2 + (3.9× 10 −6 ) 2 =1.6× 10 −5 𝑁 𝑡𝑎𝑛𝜃= 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡  𝜃= 𝑡𝑎𝑛 −1 3.9× 10 −6 1.5× 10 −5 = 14.6 𝑜 1.5 x 10-5 N q 3.9 x 10-6 N Net force