Calculate the charge per length for the capacitor

In summary, the conversation is about a student seeking help with five physics problems related to capacitors. The problems involve calculating capacitance, charge per unit length, energy stored, and other related values. The student is looking for hints and guidance to solve the problems. They have already solved one of the problems and are grateful for the help provided so far.
  • #1
hot brain
3
0

Homework Statement



I REALLY REALLY REALLY NEED YOUR HELP - I TRIED TO SOLVE THEM BUT IT'S NO USED

I HAVE 5 PROBLEMS (I NEED YOU TO HELP ME SOLVE THEM EVEN IN ONE PROBLEM )

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Q1
A cylindrical capacitor consists of a solid inner conducting core with radius 0.260 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 11.0 cm. The capacitance is 38.0 pF.

A)Calculate the outer radius of the hollow tube.B)When the capacitor is charged to 120 V, what is the charge per unit length \lambda on the capacitor?

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Q2
A parallel-plate vacuum capacitor has 8.72 J of energy stored in it. The separation between the plates is 3.60 mm. If the separation is decreased to 1.90 mm,

A)what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?

B)What is the energy now stored if the capacitor remained connected to the potential source while the separation of the plates was changed?

----------------------------------------
Q3
A 0.300-{\rm m}-long cylindrical capacitor consists of a solid conducting core with a radius of 1.15 mm and an outer hollow conducting tube with an inner radius of 2.20 mm. The two conductors are separated by air and charged to a potential difference of 6.50 V.

A)Calculate the charge per length for the capacitor.

B)Calculate the total charge on the capacitor.

C)Calculate the capacitance.

D)Calculate the energy stored in the capacitor when fully charged.

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Q4
A parallel-plate air capacitor is made by using two plates 20 cm square, spaced 4.7 {\rm mm} apart. It is connected to a 14-{\rm V} battery. If the battery remains conneated to the battery and the plates are pulled apart to a separation of 9.8 mm.

A)What is the new capacitance?

B)What is the new charge on each plate?

C)What is the new electric field between the plates?

D)What is the new energy stored in the capacitor?

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Q5
Two capacitors are connected parallel to each other and connected to the battery with voltage 30 V. Let C_1 = 9.0 \mu F,C_2 = 4.9 \mu F be their capacitances. Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite sign together.

A)By how much does the energy of the system decrease?

(I GET Q BUT WHEN WE CONNECT THEM I DON'T KNOW WHAT TO DO)
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  • #2
Q5) How much charge is there on both plates of both capacitors (give 4 numbers)

What happens if you connect a charge of +q1 coulomb with a charge of -q2 coulomb with a wire.

I don't think a lot of people will want to answer 5 problems at once, where you show no work. You probably better edit out Q1 through Q4 from this message and repost them when you have some work to show.
 
  • #3
thank you but q5 i had solve it now (U=5.71*10^-3) ^^

and i know that "^^ it will be hard to see 5 problem in one threat but i put them togther so anyone who see any easy problem of the five will answer it - and i don't want to make all the first advanced physics section page for my 5 problem

---------------------------
ONLY i need now the other 4 (i don't know what to do in these 4 i want any hit for any problem so help me ppl )
 
  • #4
hot brain said:
Q1
A cylindrical capacitor consists of a solid inner conducting core with radius 0.260 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 11.0 cm. The capacitance is 38.0 pF.

A)Calculate the outer radius of the hollow tube.


B)When the capacitor is charged to 120 V, what is the charge per unit length \lambda on the capacitor?

A. There is a formula that relates the capacitance of two plates to the area and separation of those plates. It is almost certainly in your textbook or class notes. That equation is the key here.

B. Find the charge (you don't need to solve part A first), and divide by the length. Hint: there's a basic equation used to define capacitance. Perhaps it is the very first equation that was given to you when your book or professor first started talking about capacitance.
 
  • #5
thank you for your help

i have answered all the problems in 18/4
 

Related to Calculate the charge per length for the capacitor

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It is made of two conductive plates separated by an insulating material, known as a dielectric.

2. How is the charge per length calculated for a capacitor?

The charge per length for a capacitor is calculated by dividing the total charge stored on the capacitor by the length of the plates. This value is typically measured in units of coulombs per meter.

3. What factors affect the charge per length of a capacitor?

The charge per length of a capacitor is affected by the capacitance, which is determined by the surface area of the plates, the distance between the plates, and the type of dielectric material used.

4. How is the capacitance of a capacitor determined?

The capacitance of a capacitor can be determined by using the formula C = εA/d, where C is capacitance, ε is the permittivity of the dielectric material, A is the surface area of the plates, and d is the distance between the plates.

5. Can the charge per length of a capacitor be changed?

Yes, the charge per length of a capacitor can be changed by altering the physical properties of the capacitor, such as changing the surface area or distance between the plates, or by changing the type of dielectric material used.

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