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Given E = 40xyi + 20×2 j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3).
Answer: 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.
Answer: 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.
Find the potential of V = 60sin θ/r2 at P(3,60,25)
Answer: a Explanation: V = 60sin θ/r2, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 volts.
Answer: a
See lessExplanation: V = 60sin θ/r2, put r = 3m, θ = 60 and φ = 25, V = 60 sin 60/32 = 5.774 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.
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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.
See lessA point charge 2nC is located at origin. What is the potential at (1,0,0)?
Answer: d Explanation: V = Q/(4πεr), where r = 1m V = (2 X 10-9)/(4πε x 1) = 18 volts.
Answer: d
See lessExplanation: V = Q/(4πεr), where r = 1m
V = (2 X 10-9)/(4πε x 1) = 18 volts.
Potential difference is the work done in moving a unit positive charge from one point to another in an electric field. State True/False.
Answer: a Explanation: The electric potential is the ratio of work done to the charge. Also it is the work done in moving a unit positive charge from infinity to a point in an electric field.
Answer: a
See lessExplanation: The electric potential is the ratio of work done to the charge. Also it is the work done in moving a unit positive charge from infinity to a point in an electric field.
Find the electric field intensity of transformer oil (εr = 2 approx) with density 1/4π (in 109 units)
Answer: c Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
Answer: c
Explanation: D = εE. E = (1/4π)/(2Xεo) = 4.5 X 109 units.
See lessIf the radius of a sphere is 1/(4π)m and the electric flux density is 16π units, the total flux is given by,
Answer: c Explanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ = 4π(1/16π2) X 16π = 4.
Answer: c
See lessExplanation: Total flux leaving the entire surface is, ψ = 4πr2D from Gauss law. Ψ =
4π(1/16π2) X 16π = 4.
Find the flux density of line charge of radius (cylinder is the Gaussian surface) 2m and charge density is 3.14 units?
Answer: d Explanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the radius of cylinder, which is the Gaussian surface and λ is the charge density. The density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.
Answer: d
See lessExplanation: The electric field of a line charge is given by, E = λ/(2περ), where ρ is the
radius of cylinder, which is the Gaussian surface and λ is the charge density. The
density D = εE = λ/(2πρ) = 3.14/(2π X 2) = 1/4 = 0.25.
A uniform surface charge of σ = 2 μC/m2 , is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?
Answer: b Explanation: The flux density of any field is independent of the position (point). D = σ/2 =2 X 10-6(-az)/2 = -10-6.
Answer: b
See lessExplanation: The flux density of any field is independent of the position (point). D = σ/2 =2 X 10-6(-az)/2 = -10-6.