Exploring Coulomb's Law: Electric Field & Potential Diff. of Spheres

In summary, the conversation discusses the use of Coulomb's Law and the constant k = 9 x 10^9 Nm^2/C^2 to calculate the electric field and potential difference between two metal spheres with given charges and distances. The spheres can be considered point charges for calculations. After briefly touching and separating, the charge on Sphere A can be calculated using the equation kQ1/r1 = kQ2/r2. The direction of the electric field would be towards the sphere with a charge of -6 x 10^-7 C.
  • #1
mateye10
5
0

Homework Statement


Constant in Coulomb's Law: k = 9 x 10^9 Nm^2/C^2
1) Two metal spheres are each given a charge as shown below. They are initially placed 30 cm from one another (center-to-center distance). (Qa = -6 x 10^-7 C) (Qb = -4 x 10^-7 C) (Ra = 5cm) (Rb=2cm).

A) For purposes of calculating the electric field in the region around them, the spheres can be considered the same as if they were point charges centered at the middle of the actual spheres. What is the electric field (magnitude and direction) at the midpoint between the two spheres?
B) What is the potential difference between the two spheres?
C) The spheres are briefly touched together, then separated. Now what is the charge on Sphere A?
D) Now what is the potential of Sphere B?
E) Now what is the potential difference between the two spheres?

Homework Equations


E = kq/r^2
V = kq/r

The Attempt at a Solution


A) Ea = (9 x 10^9)(-6 x 10^-7)/(.15^2) = -240,000 N/C
Eb = (9 x 10^9)(-4 x 10^-7)/(.15^2) = -160,000 N/C
Enet = (-240,000) - (-160,000) = -80,000 N/C
B) Va = (9 x 10^9)(-6 x 10^-7)/(.05) = -108,000
Vb = (9 x 10^9)(-4 x 10^-7)/(.02) = -180,000
Difference = -108,000 - -180,000 = 72,000 V
C) No idea. Something with the total charge spread over both?
D) I think I need an answer to C to do this. But still don't know.
E) Same as above. Clueless without knowing how to do C.

Thank you! I really need the help.
 
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  • #2
mateye10 said:

Homework Statement


Constant in Coulomb's Law: k = 9 x 10^9 Nm^2/C^2
1) Two metal spheres are each given a charge as shown below. They are initially placed 30 cm from one another (center-to-center distance). (Qa = -6 x 10^-7 C) (Qb = -4 x 10^-7 C) (Ra = 5cm) (Rb=2cm).

A) For purposes of calculating the electric field in the region around them, the spheres can be considered the same as if they were point charges centered at the middle of the actual spheres. What is the electric field (magnitude and direction) at the midpoint between the two spheres?
B) What is the potential difference between the two spheres?
C) The spheres are briefly touched together, then separated. Now what is the charge on Sphere A?
D) Now what is the potential of Sphere B?
E) Now what is the potential difference between the two spheres?

Homework Equations


E = kq/r^2
V = kq/r

The Attempt at a Solution


A) Ea = (9 x 10^9)(-6 x 10^-7)/(.15^2) = -240,000 N/C
Eb = (9 x 10^9)(-4 x 10^-7)/(.15^2) = -160,000 N/C
Enet = (-240,000) - (-160,000) = -80,000 N/C
What's the direction?


B) Va = (9 x 10^9)(-6 x 10^-7)/(.05) = -108,000
Vb = (9 x 10^9)(-4 x 10^-7)/(.02) = -180,000
Difference = -108,000 - -180,000 = 72,000 V
C) No idea. Something with the total charge spread over both?
D) I think I need an answer to C to do this. But still don't know.
E) Same as above. Clueless without knowing how to do C.

Thank you! I really need the help.
For (C) When they touch the spheres will both be at the same potential.
 
  • #3
The direction would be towards the sphere with the charge of -6 x 10^-7 C, right?

So if the potentials are the same I just use the equation kQ1/r1 = kQ2/r2? I know k and the radii. What would i put in for the Q's?
 
  • #4
mateye10 said:
The direction would be towards the sphere with the charge of -6 x 10^-7 C, right?

So if the potentials are the same I just use the equation kQ1/r1 = kQ2/r2? I know k and the radii. What would i put in for the Q's?
You would solve for the Q's.

BTW: You do know the sum of Q1 and Q2 , don't you ?
 
  • #5


Hello,

Thank you for sharing your attempt at solving this problem. Let me provide some guidance and corrections to your solution.

A) Your calculation for the electric field magnitude is correct. However, you forgot to include the direction. Since the charges are both negative, the electric field between them will be repulsive, meaning it will point away from the midpoint between the spheres. So, the correct answer for the electric field at the midpoint between the spheres is -80,000 N/C in the direction away from the spheres.

B) Your calculation for the potential difference is incorrect. The correct equation for potential difference is V = kq/r, not kq/r^2. Also, the potential at a point is a scalar quantity, not a vector, so you should not include the negative sign. The correct answer for the potential difference between the two spheres is 72,000 V.

C) When the spheres are touched together and then separated, the total charge on the two spheres remains the same. This is due to the law of conservation of charge. Since the total charge on the two spheres is -10 x 10^-7 C, after they are separated, the charge on Sphere A will be -10 x 10^-7 C.

D) To find the potential of Sphere B, you can use the same equation as in part B, but with the charge and distance of Sphere B instead. The correct answer for the potential of Sphere B is -45,000 V.

E) The potential difference between the two spheres will remain the same, since the total charge on the two spheres did not change. So, the potential difference between the two spheres is still 72,000 V.

I hope this helps. Keep up the good work!
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's Law related to electric fields?

Coulomb's Law is directly related to electric fields, as the force between two charged particles can be expressed in terms of the electric field created by one of the particles. The electric field is a measure of the strength of the force that would be exerted on a unit charge placed at a given point in space.

3. How can Coulomb's Law be applied to spheres?

Coulomb's Law can be applied to spheres by considering them as point charges located at their centers. The distance between the centers of the spheres is used in the equation to calculate the force between them. Additionally, the electric field and potential difference can be calculated at any point in space using the equation for a point charge.

4. What is the difference between electric field and potential difference?

Electric field and potential difference are related but distinct concepts. Electric field is a measure of the force per unit charge at a specific point in space, while potential difference is a measure of the work required to move a unit charge from one point to another. In other words, the electric field determines the force on a charge, while potential difference determines the work done on a charge.

5. Can Coulomb's Law be used to calculate the force between non-spherical objects?

Yes, Coulomb's Law can be used to calculate the force between any two charged objects, regardless of their shape. However, for non-spherical objects, the distance between the centers of the objects must be used in the equation, rather than the distance from the surface of the objects as in the case of spheres.

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