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The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)
  • a)
    12.74
  • b)
    13.47
  • c)
    12.47
  • d)
    13.74
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The potential of a coaxial cylinder with charge density 1 unit , inner...
Answer: c
Explanation: The potential of a coaxial cylinder will be ρl ln(b/a)/2πε, where ρl = 1, b = 2m and a = 1m. We get V = 12.47 X 109 volts.
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Most Upvoted Answer
The potential of a coaxial cylinder with charge density 1 unit , inner...
Given information:
- Charge density: 1 unit
- Inner radius: 1 m
- Outer radius: 2 m

To find the potential of the coaxial cylinder, we can use the following formula:

V = (1/4πε₀) * ∫(ρ/r) * dV

Where:
- V is the potential
- ε₀ is the permittivity of free space
- ρ is the charge density
- r is the distance from the center of the cylinder
- dV is the differential volume element

Let's calculate the potential step by step:

1. Find the electric field inside the inner cylinder:
Using Gauss's Law, we can find that the electric field inside the inner cylinder is given by:

E = (ρ * r) / (2ε₀)

2. Calculate the potential inside the inner cylinder:
To find the potential, we need to integrate the electric field over the distance from the center to a point inside the inner cylinder. Since the electric field is constant inside the inner cylinder, the potential can be calculated as:

V_inner = -E * Δr
= -(ρ * r) / (2ε₀) * Δr

3. Find the electric field between the inner and outer cylinders:
Inside the coaxial cylinder, the electric field is constant and given by:

E = ρ / (2ε₀)

4. Calculate the potential between the inner and outer cylinders:
To find the potential, we need to integrate the electric field over the distance from the inner to the outer cylinder. Since the electric field is constant between the cylinders, the potential can be calculated as:

V_outer = -E * Δr
= -ρ / (2ε₀) * Δr

5. Calculate the potential at the outer surface of the outer cylinder:
Since the outer surface of the outer cylinder has zero charge density, the potential at this surface is zero.

6. Calculate the potential difference between the inner surface of the inner cylinder and the outer surface of the outer cylinder:
The potential difference is given by:

ΔV = V_outer - V_inner
= -ρ / (2ε₀) * Δr - -(ρ * r) / (2ε₀) * Δr
= ρ / (2ε₀) * Δr + ρ * r / (2ε₀) * Δr
= ρ * (1 + r) / (2ε₀) * Δr

7. Substitute the given values:
To find the potential difference, we need to substitute the given values into the equation:

ΔV = 1 * (1 + 2) / (2 * 8.85 * 10^-12) * Δr
= 3 / (17.7 * 10^-12) * Δr

8. Calculate the potential:
The potential can be calculated as the negative of the potential difference:

V = -ΔV
= -3 / (17.7 * 10^-12) * Δr

9. Convert the potential into the desired units:
To convert the potential into 10^9 units, we divide the potential by 10^9:

V = -3 / (17.7 * 10^-12 * 10^9) * Δr
= -3 / 17.7 * Δr
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it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109)a)12.74b)13.47c)12.47d)13.74Correct answer is option 'C'. Can you explain this answer?
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