Need Double Check on a question

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The discussion revolves around solving a physics problem involving three charges arranged in a triangle. Key calculations include determining the force acting on charge C due to charges A and B, finding the electric field strength at a midpoint between charges B and C, and calculating the total electric potential energy of the system. Corrections were made regarding the signs of the charges and the proper formulas for potential energy and electric potential. The final consensus indicates that the total electric potential energy between charges B and C is 0 V, and further clarification is needed on the direction of the electric fields from each charge.
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Homework Statement


Given the following information:
611124_P5_U4263122123_G.gif

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!
 
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silentcoder said:

Homework Statement


Given the following information:
611124_P5_U4263122123_G.gif

Givens:

A = +2.0μC
B = +4.0 μC
C = - 4.0μC
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
You're fine up to this point

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

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
That looks like the correct magnitude.

What's the direction?
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!
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.
 
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!
 
I see you've learned to use colors on this site.
silentcoder said:
SammyS, after your suggestions I get the following answers:

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

That's correct.

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

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
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 .

4. My Attempt:
Vtotal = 9x10^9 * 4x10^-6 / 1 + 9x10^9 * -4x10^-6 / 1
Vtotal = 39600 V
No.
Check the final result.

Please let me know if these are correct, Thanks!
 
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:
silentcoder said:
Sorry, I found out that the total electric potential [STRIKE]energy[/STRIKE] midway between B and C is 0 V.

I am still confused on number 2.
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 ?
 
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