# Need Double Check on a question

1. Apr 19, 2013

### silentcoder

1. The problem statement, all variables and given/known data
Given the following information:

Givens:

A = +2.0mC
B = +4.0 mC
C = - 4.0mC
All of the sides of the triangle are 1.0 m
The inside angles are 60°

1. Determine the force acting on charge C as a result of A and B. Be sure to include a direction.
2. Determine the electric field strength and direction at a point half way between B and C. IGNORE A in this question.
3. Determine the total electric potential energy of the system of charges. Be sure to include all possible charge interactions.
4. Determine the total electric potential as a result of charges B and C at a point halfway between them. Ignore A in your thinking.

2. The attempt at a solution
1. Fb = kQq/d^2
Fb = (9x10^9) (4x10^-6)(4x10^-6) / 1^2
Fb = 0.144 N

Fa = kQq/d^2
Fa = (9x10^9) (4x10^-6) (2x10^-6)
Fa = 0.072 N

x = 0.144 + 0.072 cos 60
y = 0.072 sin 60

Therefore, resultant = 0.079 N[19 degree above horizontal]

2. Eb = kq / d^2
Eb = (9x10^9)(4x10^-6) / 0.5^2
Eb = 144000 N/C

Ec = kq / d^2
Ec = (9x10^9)(4x10^-6) / 0.5^2
Ec = 144000 N/C

E = Eb + Ec
E = 144000 + 144000
E = 288000 N/C

Therefore, electric field strength is 2.9x10^5 N/C

3. Pe = k(qa qb / rab + qa qc / rac + qb qc / rbc)
Pe = (9x10^9)(2x10^-6 * 4x10^-6 / 1^2 + 2x10^-6 * 4x10^-6 / 1^2 + 4x10^-6 * 4x10^-6 / 1^2)
Pe = (9x10^9) (8x10^-12 + 8x10^-12 + 1.6x10^-11)
Pe = 9.216 x 10^-24 J

Therefore, total electric potential energy of the system of charges is 9.2 x 10^-24 J

4. Needs help solving this question.

Thanks!

2. Apr 19, 2013

### SammyS

Staff Emeritus
You're fine up to this point

Make sure your calculator is in degree mode for the following, which isn't correct.

That looks like the correct magnitude.

What's the direction?
I assume those charges are in units of μC .

For part (3) -- Two mistakes.

You ignored the signs of the charges.

You should divide by the distance between charges, not the square of the distance.

For part (4): This is similar to part(2), except you're to find electric potential at the mid-point, rather than finding electric field.

3. Apr 19, 2013

### silentcoder

SammyS, after your suggestions I get the following answers:

1. Resultant = 0.19 N[19 degrees above horizontal]

2. Electric Field Strength: 2.9x10^2 N/C [WHAT WILL BE THE DIRECTION?]

3. Is this correct?
Pe = k(qa qb / rab + qa qc / rac + qb qc / rbc)
Pe = (9x10^9)(2x10^-6 * 4x10^-6 / 1 + 2x10^-6 * -4x10^-6 / 1 + 4x10^-6 * -4x10^-6 / 1)
Pe = (9x10^9) (8x10^-12 + -8x10^-12 + -1.6x10^-11)
Pe = -0.144 N/C

4. My Attempt:
Vtotal = 9x10^9 * 4x10^-6 / 1 + 9x10^9 * -4x10^-6 / 1
Vtotal = 39600 V

Please let me know if these are correct, Thanks!

4. Apr 19, 2013

### SammyS

Staff Emeritus
I see you've learned to use colors on this site.
That's correct.

I asked you the direction first. How can you add the electric field from each charge, without knowing the direction for the field from each?

The above is correct.

Notice that the potential energy due to the interaction of charges A & B cancels with the potential energy due to the interaction of charges A & C .

No.
Check the final result.

5. Apr 19, 2013

### silentcoder

Sorry, I found out that the total potential energy between B and C is 0 V.

I am still confused on number 2.

Last edited: Apr 19, 2013
6. Apr 20, 2013

### SammyS

Staff Emeritus
Those corrections above are important. You're finding electric potential, not finding potential energy.

Regarding part (2):

What is the direction of the electric field due to charge B ?

What is the direction of the electric field due to charge C ?