# Homework Help: Another point charge question (electric field)

1. Jun 4, 2012

### curiousjoe94

1. The problem statement, all variables and given/known data

A +15μC point charge Q1 is at a distance of 30mm from -30μC charge Q2.

Show that the electric potential is zero at a point between the two charges which is 10mm from Q1 and 20mm from Q2.

3. The attempt at a solution

I know that V = Ed (where V = electric potential, E = electric field strength, d = distance from the point to the charge)

I have two problems though:

1) I'm not sure where electric potential is being measure from, from charge Q1 or Q2, so I don't know what value (10mm or 20mm) to take as distance, d. With the examples I'm given in the textbook, its simple - just one charge point and a test charge.

2) I know I'm not supposed to find 'E', the resultant electric field strength, because its only in the next question that I'm asked to calculate this. So I think this makes the equation I've given at the top ( V = Ed) redundant as it probably won't be used.

so I'm kinda stuck on where to start.

2. Jun 4, 2012

### Infinitum

Think that you weren't given 10mm and 20mm, and let the distance of zero potential from the charge 15μC be d. So the other distance is....?

You only have to use the potential formula you've written, you need to see where the sum of potentials is zero. Remember, potential is a scalar quantity!

3. Jun 4, 2012

### Simon Bridge

1. electric potential is being measured from an infinite distance away ... but you don't care about that so much because you can see that the charges have opposite signs so the potential from one will be negative and the potential from the other will be positive and at some place they will sum to zero.

You will probably get away with noticing that Q2 = -2Q1 and reasoning what that means for the ratio of distances for the potentials to be the same (but opposite). Some people find it easier to think about if they start out with the general formula for the electric field and derive the formula for the potential from that.

2. you only need to compute the electric field second - you can still use the formula if you want to derive new ones.

When you are stuck for were to start it can help to collect everything you know.

4. Jun 4, 2012

### curiousjoe94

So for Q1:

v = Ed

Q2:

v = E(30 - d)

summing them up gives:

Ed + E30 - Ed = 2v

E30 = 2v

^ well, I've definitely done something wrong there.

5. Jun 4, 2012

### curiousjoe94

Yeah I see how that works, but I'm still not quite sure what to do if the charges and distances aren't so conviniently set up like this.

6. Jun 4, 2012

### curiousjoe94

quick question, if the charge Q is negative, would the electric potential value be decreasing the closer you get to Q?

whereas with a positive charge, electric potential increases further inwards towards it, right?

7. Jun 4, 2012

### Villyer

Mark your E's; one refers to the 15μC charge and the other refers to the 30μC, right?

8. Jun 4, 2012

### curiousjoe94

Ok so it would be:

E'd + E30 - Ed = 2v

where E' refers to the 15μC charge and E refers to 30μC charge, still not sure where to go from here.

9. Jun 4, 2012

### Simon Bridge

If it is not conveniently set up you can still work the ratios - it just won't be a simple ratio.

OR:
derive the formula for the potential for some arbitrary position x and solve for the position that makes the potential zero. It usually helps to define x=0 to be on one of the charges.

Note: you know the equation for E, and you know how E and V are related - so you can construct an equation for V.

10. Jun 4, 2012

### Infinitum

You can make equations using d, and writing out what E is for each. This works for all cases. Example,

$$V_1 = E_1d = \frac{kQ_1}{d^2}\cdot d$$

Similarly for the other charge...