# Calculate the maximum voltage that can be applied

• xxwwyytt
In summary: The question is asking for the maximum voltage that can be applied without breakdown occurring, not the maximum charge that can be stored. In summary, the conversation discusses the use of a Geiger-Muller counter, a detector of radiation, and the calculation of the maximum voltage that can be applied between the central wire and the conducting cylinder before breakdown occurs in the gas. The conversation mentions the use of Gauss' Law and the dielectric strength of the gas to calculate the maximum voltage. It is determined that the equation should be solved for Qmax first before plugging in the values for the voltage.
xxwwyytt

## Homework Statement

A detector of radiation called a Geiger-Muller counter consists of a closed, hollow, conducting cylinder with a fine wire along its axis. Suppose that the internal diameter of the cylinder is 3.00 cm and that the wire along the axis has a diameter of 0.2 mm. If the dielectric strength of the gas between the central wire and the cylinder is 1.00 ✕ 106 V/m, calculate the maximum voltage that can be applied between the wire and the cylinder before breakdown occurs in the gas.

E=-∫Vds
∫EdA = Qenc/ε0

## The Attempt at a Solution

Since the electric field is not constant between the cylinder and the wire, I tries to derive for the electric field using Gauss' law from the wire to the cylinder using radius 1.5e-2m and 0.1e-3m. However the equation comes out like Q/(2πlε0)ln(r). l is for the length of the gaussian surface which i don't know and i don't see anywhere that I can cancel l . Besides, I have no idea where I can use that dielectric strength since electric field vary.

xxwwyytt said:
Q/(2πlε0)ln(r). l is for the length of the gaussian surface which i don't know and i don't see anywhere that I can cancel l
You are asked for the maximum voltage, not the maximum charge. The longer the tube and wire, the greater the total charge.
Work with the charge per unit length. Relate that to the voltage and to the field (as function of r).

haruspex said:
You are asked for the maximum voltage, not the maximum charge. The longer the tube and wire, the greater the total charge.
Work with the charge per unit length. Relate that to the voltage and to the field (as function of r).
I tried to relate to the E field by using Gauss's Law, and I get E = λ/2πrε0. Integrate E, I get V= (λ/2πε0)ln(r), and then I don't know where I should integrate from. Should I integrate from 1.5e-2m to 0.1e-3m or do I get r by plugging the dielectric strength into the equation of E?

xxwwyytt said:
Should I integrate from 1.5e-2m to 0.1e-3m
If you do that, what exactly does the resulting equation tell you?
xxwwyytt said:
do I get r by plugging the dielectric strength into the equation of E?
If you do that, what exactly does the resulting equation tell you?

haruspex said:
If you do that, what exactly does the resulting equation tell you?

If you do that, what exactly does the resulting equation tell you?
Thanks! I figure out that I should get Qmax first and then plug in 1.5e-2 and 0.1e-3 to find the Vmax.

xxwwyytt said:
Thanks! I figure out that I should get Qmax first and then plug in 1.5e-2 and 0.1e-3 to find the Vmax.
Perhaps, but how do you plan to find Qmax?

## 1. What is the maximum voltage that can be applied?

The maximum voltage that can be applied is the highest amount of electric potential that can be used in a circuit without causing damage or malfunction.

## 2. How is the maximum voltage calculated?

The maximum voltage is calculated by using Ohm's Law, which states that the maximum voltage is equal to the current multiplied by the resistance in a circuit.

## 3. Can the maximum voltage be exceeded?

Exceeding the maximum voltage can be dangerous and can cause damage to the circuit or electrical equipment. It is important to always stay within the maximum voltage limit.

## 4. What factors affect the maximum voltage that can be applied?

The maximum voltage that can be applied is affected by the resistance of the circuit, the type of materials used, and the voltage rating of the components.

## 5. Why is it important to calculate the maximum voltage in a circuit?

Calculating the maximum voltage in a circuit is important to ensure the safety and proper functioning of the circuit and its components. It also helps prevent damage to electrical equipment and potential hazards such as electric shock or fire.

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