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Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 3m
b Explanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m. By Gauss law, ψ = Q D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units
b
See lessExplanation: The radius is 3m, hence it will enclose one Gaussian cylinder of R = 2m.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 3) = σ(2π X 2), Thus D = 10/3 units
Three charged cylindrical sheets are present in three spaces with σ = 5 at R = 2m, σ = -2 at R = 4m and σ = -3 at R = 5m. Find the flux density at R = 1m.
a Explanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero
a
See lessExplanation: Since 1m does not enclose any cylinder (three Gaussian surfaces of
radius 2m, 4m, 5m exists), the charge density and charge becomes zero according
to Gauss law. Thus flux density is also zero
Find the potential due to a charged ring of density 2 units with radius 2m and the point at which potential is measured is at a distance of 1m from the ring
d Explanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r = 1m. We get V = 72π volts
d
See lessExplanation: The potential due to a charged ring is given by λa/2εr, where a = 2m and r
= 1m. We get V = 72π volts
The potential of a coaxial cylinder with charge density 1 unit , inner radius 1m and outer cylinder 2m is (in 109 )
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.
A circular disc of radius 5m with a surface charge density ρs = 10sinφ is enclosed by surface. What is the net flux crossing the surface?
The total charge of a surface with densities 1,2,…,10 is
The total charge of a surface with densities 1,2,…,10 is
See lessWhat is the potential difference between 10sinθcosφ/r 2 at A(1,30,20) and B(4,90,60)?
c Explanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb = 10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 – 0.3125 = 4.386 volts
c
See lessExplanation: Potential at A, Va = 10sin30cos20/12 = 4.6985 and Potential at B, Vb =
10sin90cos60/42 = 0.3125. Potential difference between A and B is, Vab = 4.6985 –
0.3125 = 4.386 volts
Given E = 40xyi + 20×2 j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3)
: b Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get Vpq on integrating from Q to P. Vpq = 106 volts
: b
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get
Vpq on integrating from Q to P. Vpq = 106 volts
A point charge 0.4nC is located at (2, 3, 3). Find the potential differences between (2, 3, 3)m and (-2, 3, 3)m due to the charge
c Explanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA = 1m and rB = 4m. Thus Vab = 2.7 volts.
c
See lessExplanation: Vab = (Q/4πεo)(1/rA) + (1/rB), where rA and rB are position vectors rA =
1m and rB = 4m. Thus Vab = 2.7 volts.
Six equal point charges Q = 10nC are located at 2,3,4,5,6,7m. Find the potential at origin.
d Explanation: V = (1/4πεo) ∑Q/r = (10 X 10-9 /4πεo) (0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.
d
See lessExplanation: V = (1/4πεo) ∑Q/r = (10 X 10-9
/4πεo)
(0.5 + 0.33 + 0.25 + 0.2 + 0.166 + 0.142) = 143.35 volts.
Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109 units)
c Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units
c
See lessExplanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units