Two conducting spheres are initially not connected. Sphere A has a charge of +2q

Question

Two conducting spheres are initially not connected. Sphere A has a charge of +2q and radius r_0, Sphere B has a charge of +4q and radius 2r_0. a. Find the surface charge 77 for each sphere. Which one has the larger charge per unit area? b. Now the two spheres are connected by a very thin conducing wire. What is the charge per area on each sphere? What is the total charge on each sphere? You may assume the surface area of the wire is so small compared to the spheres that it can be ignored. c. What is the electric field at the surface of each sphere? d. What is the electric potential at the surface of each sphere? Did you calculate the answer you expected? Explain.

Explanation / Answer

a.

the surface charge density A for smaller sphere is,

A = 2q/(4r02)

the surface charge density B for larger sphere is,

B = 4q/[4(2r0)2]

or, B= q/(4r02)

So, the sphere A has larger surface charge per unit area.

***************************************************************************************************

b.

When the two spheres are connected, their potential will be equal and the charges redistribution will take place while sum of total charge on the two spheres remains the same.

Let charge on A be q1 and charge on B be q2. Since their potential are equal, we have

q1/(4r0) = q2/[4(2r0)]

or, q1 = q2/2

also q1 + q2 = 6q

so, from above two equations, we have,

q1 = 2q, an q2 = 4q

charges on A and B remain same as before.

So charge per unit area will also remain the same as calculated in a. above.

***************************************************************************************************

c.

Electric field at A is,

EA = 2q/(4r02)

Electric field at B is,

EB = 4q/(4(2r02)

or, EB = q/(4r02)

***************************************************************************************************

d.

Potential at A is,

VA = 2q/(4r0)

Potential at B is,

VB = 4q/(42r0)

or, VB = 2q/(4r0)

***************************************************************************************************
This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.