Bartleby Related Questions Icon

Related questions

Question
Can you help me figure out what I did wrong on this test question?
[画像:1. (50 points) A proton is initially placed at rest a distance d = 3.0 mm away from the center of a uniformly-charged disk of radius r = 12.0 cm carrying -46 nC of charge, as shown in the diagram immediately below (not to scale): a) (15 points) What is the initial force on the proton? (Don't forget that force is a vector!) surface of uniform charge luf to diff in distance and redi Using E on JE =D A= (12m) = .045m2 a=" -6 E=26 = -1.07x1070/2 1.885 ×ばつ 10-12 (1/Nm2) 1- -460C .045m 2 a = -1016.8 nc/m2 =-1.07x10-6 C "C/M2 =-60451 N/C F=qE=1.6x10-19 - 60451N/C Now, a second identical disk carrying +46 nC of charge is placed 6.0 mm above the first disk, as shown in the new diagram below. A proton is placed at rest between the disks' centers at a distance d = 3.0 mm from each disk, as shown in the diagram below (not to scale): +46 F9.7x10 1/2 -46 d d b) (15 points) What is the new force on the proton? c) (10 points) In this new two-plate situation, what is the electric potential at the proton's initial position? You may assume V = 0 at the negative plate. d) (10 points) What is the proton's initial electric potential energy? b. Since & is opposite, it will push the proton. Charge is equal in magnitude, and d and dick are the Some so FI will be the Same. Frew=27 Fret = 0 N-10 C. V=Es AV= 60451 W/C -2 V= 003 m. 362√ = 1812 d. =qv .000m 006m = 362V v=AV 1815 v- 1.6x10-19 C = 2.9x 10-17-1/2 \-]
expand button
Transcribed Image Text:1. (50 points) A proton is initially placed at rest a distance d = 3.0 mm away from the center of a uniformly-charged disk of radius r = 12.0 cm carrying -46 nC of charge, as shown in the diagram immediately below (not to scale): a) (15 points) What is the initial force on the proton? (Don't forget that force is a vector!) surface of uniform charge luf to diff in distance and redi Using E on JE =D A= (12m) = .045m2 a=" -6 E=26 = -1.07x1070/2 1.885 ×ばつ 10-12 (1/Nm2) 1- -460C .045m 2 a = -1016.8 nc/m2 =-1.07x10-6 C "C/M2 =-60451 N/C F=qE=1.6x10-19 - 60451N/C Now, a second identical disk carrying +46 nC of charge is placed 6.0 mm above the first disk, as shown in the new diagram below. A proton is placed at rest between the disks' centers at a distance d = 3.0 mm from each disk, as shown in the diagram below (not to scale): +46 F9.7x10 1/2 -46 d d b) (15 points) What is the new force on the proton? c) (10 points) In this new two-plate situation, what is the electric potential at the proton's initial position? You may assume V = 0 at the negative plate. d) (10 points) What is the proton's initial electric potential energy? b. Since & is opposite, it will push the proton. Charge is equal in magnitude, and d and dick are the Some so FI will be the Same. Frew=27 Fret = 0 N-10 C. V=Es AV= 60451 W/C -2 V= 003 m. 362√ = 1812 d. =qv .000m 006m = 362V v=AV 1815 v- 1.6x10-19 C = 2.9x 10-17-1/2 \-
[画像:1. (50 points) A proton is initially placed at rest a distance d = 3.0 mm away from the center of a uniformly-charged disk of radius r = 12.0 cm carrying -46 nC of charge, as shown in the diagram immediately below (not to scale): a) (15 points) What is the initial force on the proton? (Don't forget that force is a vector!) surface of uniform charge luf to diff in distance and redi Using E on JE =D A= (12m) = .045m2 a=" -6 E=26 = -1.07x1070/2 1.885 ×ばつ 10-12 (1/Nm2) 1- -460C .045m 2 a = -1016.8 nc/m2 =-1.07x10-6 C "C/M2 =-60451 N/C F=qE=1.6x10-19 - 60451N/C Now, a second identical disk carrying +46 nC of charge is placed 6.0 mm above the first disk, as shown in the new diagram below. A proton is placed at rest between the disks' centers at a distance d = 3.0 mm from each disk, as shown in the diagram below (not to scale): +46 F9.7x10 1/2 -46 d d b) (15 points) What is the new force on the proton? c) (10 points) In this new two-plate situation, what is the electric potential at the proton's initial position? You may assume V = 0 at the negative plate. d) (10 points) What is the proton's initial electric potential energy? b. Since & is opposite, it will push the proton. Charge is equal in magnitude, and d and dick are the Some so FI will be the Same. Frew=27 Fret = 0 N-10 C. V=Es AV= 60451 W/C -2 V= 003 m. 362√ = 1812 d. =qv .000m 006m = 362V v=AV 1815 v- 1.6x10-19 C = 2.9x 10-17-1/2 \-]
expand button
Transcribed Image Text:1. (50 points) A proton is initially placed at rest a distance d = 3.0 mm away from the center of a uniformly-charged disk of radius r = 12.0 cm carrying -46 nC of charge, as shown in the diagram immediately below (not to scale): a) (15 points) What is the initial force on the proton? (Don't forget that force is a vector!) surface of uniform charge luf to diff in distance and redi Using E on JE =D A= (12m) = .045m2 a=" -6 E=26 = -1.07x1070/2 1.885 ×ばつ 10-12 (1/Nm2) 1- -460C .045m 2 a = -1016.8 nc/m2 =-1.07x10-6 C "C/M2 =-60451 N/C F=qE=1.6x10-19 - 60451N/C Now, a second identical disk carrying +46 nC of charge is placed 6.0 mm above the first disk, as shown in the new diagram below. A proton is placed at rest between the disks' centers at a distance d = 3.0 mm from each disk, as shown in the diagram below (not to scale): +46 F9.7x10 1/2 -46 d d b) (15 points) What is the new force on the proton? c) (10 points) In this new two-plate situation, what is the electric potential at the proton's initial position? You may assume V = 0 at the negative plate. d) (10 points) What is the proton's initial electric potential energy? b. Since & is opposite, it will push the proton. Charge is equal in magnitude, and d and dick are the Some so FI will be the Same. Frew=27 Fret = 0 N-10 C. V=Es AV= 60451 W/C -2 V= 003 m. 362√ = 1812 d. =qv .000m 006m = 362V v=AV 1815 v- 1.6x10-19 C = 2.9x 10-17-1/2 \-
Expert Solution
Check Mark