Need Double Check on a question

[silentcoder]

Homework Statement

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!

 

[SammyS]

Homework Statement

Given the following information:

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.

 

[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!

 

[SammyS]

I see you've learned to use colors on this site.

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!

 

[silentcoder]

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

I am still confused on number 2.

 

[SammyS]

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 ?