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seiferseph
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I have some more questions from a physics worksheet, this time on electrical forces. here are the questions and the answers i got, thanks in advance!
Questions:
http://i2.photobucket.com/albums/y15/seiferseph/electircalforces.jpg
Answers:
1) Fe = kQq/R^2, i get 3.6 x 10^10 N
2) a) Fg = GMm/R^2, i get 7.80 x 10^-47 N
b) Fe = kQq/R^2, using 1.6x10^-19 for both Q and -1.6x10^-19 for q, i get Fe = -8.86x10^-8 N
3) Q = n*e, Q/e = n. n = 2.5x10^19. i have a question though, can it be negative? (the charge is negative)
4) How do i do this one? is it just a simple ratio problem?
5) a) i calculated the force of each one seperately, i used 2*1.6x10^-19 for the charge of the middle one, because it has two protons.
F 1 on middle = k*q1*qmiddle/R^2 = 3.2x10-16
F 2 on middle = k*q2*qmiddle/R^2 = 1.44x10-15
and i get 1.76x10^15 N to the right
b) F = m*a, F/m = a, where m is 4*1.67x10^-27 = 2.6x10^11 m/s^2
thanks!
Questions:
http://i2.photobucket.com/albums/y15/seiferseph/electircalforces.jpg
Answers:
1) Fe = kQq/R^2, i get 3.6 x 10^10 N
2) a) Fg = GMm/R^2, i get 7.80 x 10^-47 N
b) Fe = kQq/R^2, using 1.6x10^-19 for both Q and -1.6x10^-19 for q, i get Fe = -8.86x10^-8 N
3) Q = n*e, Q/e = n. n = 2.5x10^19. i have a question though, can it be negative? (the charge is negative)
4) How do i do this one? is it just a simple ratio problem?
5) a) i calculated the force of each one seperately, i used 2*1.6x10^-19 for the charge of the middle one, because it has two protons.
F 1 on middle = k*q1*qmiddle/R^2 = 3.2x10-16
F 2 on middle = k*q2*qmiddle/R^2 = 1.44x10-15
and i get 1.76x10^15 N to the right
b) F = m*a, F/m = a, where m is 4*1.67x10^-27 = 2.6x10^11 m/s^2
thanks!
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