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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

2) a) Fg = GMm/R^2, i get

b) Fe = kQq/R^2, using 1.6x10^-19 for both Q and -1.6x10^-19 for q, i get Fe =

3) Q = n*e, Q/e = n. n =

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

b) F = m*a, F/m = a, where m is 4*1.67x10^-27 =

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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