Point at which Electric Field = 0

[deleted388502]

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Hi! I was wondering if someone could explain to me how they intuitively reason out these sorts of problems.
I know that it obviously had to be either the first or second due to size of the charge but I was confused as to how to determine which one it would be without writing out the equations.

Thanks!

 

[Cawolf]

I would just think of how the field decays with 1/r^2.

I mean writing out the equation would give you a real simple solution but you can see that 1/r^2 = 16/r2^2

It works out that the larger charge needs to be sqrt(q1/q2) distance farther - always.

So here the larger charge is 16x larger - so it needs to be 4 times farther than the other charge (assuming opposite charges).

You saw how it was easy to rule out the last 2 choices.

When you look at the 2nd choice you can see that it would be 5x farther, leaving choice 1 as the answer.

 

[sillyjoe]

I would just think of how the field decays with 1/r^2.

I mean writing out the equation would give you a real simple solution but you can see that 1/r^2 = 16/r2^2

It works out that the larger charge needs to be sqrt(q1/q2) distance farther - always.

So here the larger charge is 16x larger - so it needs to be 4 times farther than the other charge (assuming opposite charges).

You saw how it was easy to rule out the last 2 choices.

When you look at the 2nd choice you can see that it would be 5x farther, leaving choice 1 as the answer.

Is it necessary to do the math? Since it is multiple choice, can you just assume that the answer will be whichever is farthest to the left of the -1uC charge?

 

[Cawolf]

No I don't think it will always be what is farthest to the left.

If the positive charge was greater, say +25 microC.

Then 6mm to the left (choice 2) would be correct.

That way the greater charge would be 5x farther and the fields would cancel - note it is not the farthest left.

 

[sillyjoe]

No I don't think it will always be what is farthest to the left.

If the positive charge was greater, say +25 nC.

Then 6mm to the left (choice 2) would be correct.

That way the greater charge would be 5x farther and the fields would cancel - note it is not the farthest left.

Good stuff. Thanks! Sorry if I hijacked your thread @avenlea

 

[Cawolf]

I think that it was a great addition @sillyjoe !

 

[deleted388502]

I would just think of how the field decays with 1/r^2.

I mean writing out the equation would give you a real simple solution but you can see that 1/r^2 = 16/r2^2

It works out that the larger charge needs to be sqrt(q1/q2) distance farther - always.

So here the larger charge is 16x larger - so it needs to be 4 times farther than the other charge (assuming opposite charges).

You saw how it was easy to rule out the last 2 choices.

When you look at the 2nd choice you can see that it would be 5x farther, leaving choice 1 as the answer.

Okay so just to verify-

if we use 8mm, we're looking at the charges being 32mm and 8mm away from this new charge, and with the 6mm charge being 30mm and 6mm charge, but we want them to be 6mm because we want a 4x charge separation because their charge differential is only 4x, and because of that, if their charge separation is larger, the electric field will not be equal between them?

I stared at this intermittently throughout the day yesterday and it took me a while to actually ~see~ it and if I saw this on my actual MCAT I don't think I would've been able to figure it out, lol, how did you guys approach this - just knowing the equation and understanding the field relationship??

And no problem @sillyjoe your input is always welcome!

 

[sillyjoe]

Okay so just to verify-

if we use 8mm, we're looking at the charges being 32mm and 8mm away from this new charge, and with the 6mm charge being 30mm and 6mm charge, but we want them to be 6mm because we want a 4x charge separation because their charge differential is only 4x, and because of that, if their charge separation is larger, the electric field will not be equal between them?

I stared at this intermittently throughout the day yesterday and it took me a while to actually ~see~ it and if I saw this on my actual MCAT I don't think I would've been able to figure it out, lol, how did you guys approach this - just knowing the equation and understanding the field relationship??

Let's just do it strictly mathematically and then when you do it out yourself you can start conceptualizing it.

Let's make:
Q1 = -1 charge and
Q2 = +16 charge
r = distance from Q1
r' = distance from Q2

The equation we use for electric fields are F = qE or E = F/q.

We also know that, F = kQ1Q2/r^2. If we substitute this equation for F in the electric field equation we get:

E = (kQ1Q2/r^2) / Q which is equal to E = kq / r^2 . Now we have a workable equation for this problem

We want a point where the electric field from the POSITIVE charge and NEGATIVE charge is equal so they cancel each other out, so:

E(negative charge) = E(positive charge)

kQ1 / r^2 = kQ2/ r'^2

We can get rid of the constant because f*** constants. We can also just use absolute values for this:

1 / r^2 = 16 / r'^2

16r^2 = r'^2

r'^2 / r^2 = 16

r' / r = 4

This is just saying the ratio of r':r is 4:1

So r' (distance from the 16uC charge) must be 4 times as big as r (distance from 1 uC charge).

 

[sillyjoe]

And no problem @sillyjoe your input is always welcome!

 

[deleted388502]

Cool beans. Thanks @Cawolf as well!