Recent content by coz

I'm sorry to ask another question, but I think I'm missing something here. For the horizontal component I get Ex = 0 Ey = -4.5x104 NC-1 For the vertical component I get Ey = 0 Ex = 6.25x104 NC-1 To calculate the net force on the oil drop would it be better to use Felectric =...

Can I just check something. Once you have worked out the horizontal and vertical components you have a value for Ex and Ey. To then work out the net force on the oil drop could you use: Fnet = (Ex + Ey) x q Thanks

Thank you so much for all your help :)

Cool, thank you so much. I get r = 1.15x10^-14 m Just one more thing, the question says to express your answer in terms of the radius of the zinc nucleus. I'm not sure what this means. The radius of the zinc nucleus is 2.00x10^-14 m. Thanks again :)

The initial kinetic energy is 1.20x10^-12 J, so the electric energy is also 1.20x10^-12 J. Can I rearrange E=q1q2/4\pi\epsilon0r to find r?

Sorry, I meant electrostatic potential energy when I said E_el

Sorry for rambling! m=F/g = 27.486kg This seems too simple.

The only equations I know for gravitational force are; Fgrav = -Gm1m2/r2 Fgrav (on m at r) = mg(r) But both of these need the mass to work out the gravitational force. I could use F=ma to work out the mass, but this seems too easy! Also you mentioned torque earlier, so I don't think this...

Ok, so the charges are like and so they repel, and the RHS repels with greater force than the LHS, because the charge is higher there.

Not yet, but I won't let it beat me! :)

Obviously I didn't use the tags correctly! How do they work then? Thanks again!

Hi, no I am not the same person as cerium, just have a similar question to work out. I am new to this, so thanks for the tips! For the LHS I have [F][/el] = [8.988x10][/9] N m2 C-2 x ([5.0x10][/-6] x [5.0x10][/-6])/(0.1m)2 = 22.47N For the RHS I have [F][/el] = [8.988x10][/9] N m2 C-2...

Hi tiny-tim, Can I just check, to find the electrostatic force on either side of the see saw would you use the equation: F_el = 1/4pi\epsilon_0 x q_1xq_2/r_2

I don't think my above thoughts would work, as E_el would only equal zero when r = infinity and I am trying to work out r (in terms of the radius of the zinc nucleus). Back to the drawing board! :)

Does this mean you can assume E_el is zero and rearrange the equation E_el = q_1xq_2/4piE_or so that r is the subject and work out the answer that way? Thanks