Prelim Physics Help (electric fields) (1 Viewer)

[turntaker]

hi guys,
Is anyone able to help with this question : )

I am not exactly sure how to do it.

Thanks in advance

 

[anomalousdecay]

I think this is all coulomb's law. So:



a) Increase separation, decrease force on each by inverse square law.

So for distance r, we have F.

Distance 2r, distance is doubled, so we get 1/2^2 = 1/4, so the force is F/4

b) Same thing, but we get 9F.

c) Charge is doubled for both. So you get double from one, so 2F, then double from the other, another 2F.

As a result the total force is 4F.

d) This time we have half on one, quarter on the other. So we get F/(4x2) = F/8 as the new force.

e) So here the ratio is F (2 x 0.25 x 1/9)



12) use coulomb's law.

 

[turntaker]

I think this is all coulomb's law. So:



a) Increase separation, decrease force on each by inverse square law.

So for distance r, we have F.

Distance 2r, distance is doubled, so we get 1/2^2 = 1/4, so the force is F/4

b) Same thing, but we get 9F.

c) Charge is doubled for both. So you get double from one, so 2F, then double from the other, another 2F.

As a result the total force is 4F.

d) This time we have half on one, quarter on the other. So we get F/(4x2) = F/8 as the new force.

e) So here the ratio is F (2 x 0.25 x 1/9)



12) use coulomb's law.

thanks man, I'll do it properly now.
So I don't need to worry about K or q1 and q2?

 

[turntaker]

so F is inversely proportional to r?

 

[anomalousdecay]

thanks man, I'll do it properly now.
So I don't need to worry about K or q1 and q2?

K no. K is the constant.

However q1 and q2 will determine the actual force. Here we can just use the formula as ratios

So if we get double q1 = 2q1, then in the formula yuo will get a ratio which can be compared to F, hence giving you the final result in the end.

so F is inversely proportional to r?

F is inversely proportional to r squared.

F is also proportional to q1 and q2.