What is the electric field and potential difference between two charged spheres?

AI Thread Summary
The discussion revolves around calculating the electric field and potential difference between two charged metal spheres. The electric field at the midpoint is found to be -80,000 N/C, indicating the direction is towards the more negatively charged sphere. The potential difference between the spheres is calculated to be 72,000 V. When the spheres are touched together, they will share their total charge, leading to a need for further calculations to determine the new charges and potentials. The participants express confusion about the charge distribution after touching and how to proceed with the calculations.
mateye10
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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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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.
 
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?
 
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 ?
 
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