432. The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109 )
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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.
c
See lessExplanation: 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.
2m and a = 1m. We get V = 12.47 X 109 volts.
2m and a = 1m. We get V = 12.47 X 109 volts.
See lessC 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 X109volts.
C
See lessExplanation: 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 X109volts.
Answer: 12.47 X 10⁹volts. 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 10⁹volts.
Answer: 12.47 X 10⁹volts.
See lessExplanation: 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 10⁹volts.
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.
Answer: c
See lessExplanation: 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.
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.
c)
See lessExplanation: 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.