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

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field and potential difference between two charged metal spheres, Qa = -6 x 10^-7 C and Qb = -4 x 10^-7 C, positioned 30 cm apart. The electric field at the midpoint is calculated to be -80,000 N/C, indicating the direction towards the more negatively charged sphere. The potential difference between the spheres is determined to be 72,000 V. Additionally, when the spheres are touched together, they will share their total charge, which is crucial for determining the new charge on Sphere A and the potential of Sphere B.

PREREQUISITES
  • Coulomb's Law (k = 9 x 10^9 Nm^2/C^2)
  • Electric field calculation (E = kq/r^2)
  • Electric potential calculation (V = kq/r)
  • Understanding of charge conservation principles
NEXT STEPS
  • Learn about charge distribution when conductive objects are brought into contact.
  • Study the concept of electric field lines and their representation.
  • Explore the relationship between electric potential and electric field.
  • Investigate the effects of varying distances on electric field strength and potential difference.
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields and potentials in charged systems.

mateye10
Messages
5
Reaction score
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.
 
Physics news on Phys.org
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 ?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
23
Views
4K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K