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Evaluate Gauss law for D = 5r2 /4 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.
Answer: c Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated. Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.
Answer: c
See lessExplanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated.
Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.
Coulomb’s law can be derived from Gauss law. State True/ False
Answer: a Explanation: Gauss law, Q = ∫∫D.ds By considering area of a sphere, ds = r2sin θ dθ dφ. On integrating, we get Q = 4πr2D and D = εE, where E = F/Q. Thus, we get Coulomb’s law F = Q1 x Q2/4∏εR2.
Answer: a
Explanation: Gauss law, Q = ∫∫D.ds
By considering area of a sphere, ds = r2sin θ dθ dφ.
On integrating, we get Q = 4πr2D and D = εE, where E = F/Q.
See lessThus, we get Coulomb’s law F = Q1 x Q2/4∏εR2.
Surface integral is used to compute
Answer: b Explanation: Surface integral is used to compute area, which is the product of two quantities length and breadth. Thus it is two dimensional integral
Answer: b
See lessExplanation: Surface integral is used to compute area, which is the product of two
quantities length and breadth. Thus it is two dimensional integral
Gauss law for electric field uses surface integral. State True/False
Answer: a Explanation: Gauss law states that the electric flux passing through any closed surface is equal to the total charge enclosed by the surface. Thus the charge is defined as a surface integral.
Answer: a
See lessExplanation: Gauss law states that the electric flux passing through any closed surface
is equal to the total charge enclosed by the surface. Thus the charge is defined as a
surface integral.
The energy stored in the inductor 100mH with a current of 2A is
Answer: a Explanation: dw = ei dt = Li di, W = L∫ i.di Energy E = 0.5LI2 = 0.5 X 0.1 X 22 = 0.2 Joule.
Answer: a
See lessExplanation: dw = ei dt = Li di, W = L∫ i.di
Energy E = 0.5LI2 = 0.5 X 0.1 X 22 = 0.2 Joule.
Line integral is used to calculate
Answer: d Explanation: Length is a linear quantity, whereas area is two dimensional and volume is three dimensional. Thus single or line integral can be used to find length in general.
Answer: d
See lessExplanation: Length is a linear quantity, whereas area is two dimensional and volume is three dimensional. Thus single or line integral can be used to find length in general.
The potential in a lamellar field is
Answer: b Explanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.
Answer: b
See lessExplanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.
A field in which a test charge around any closed surface in static path is zero is called
Answer: d Explanation: Work done in moving a charge in a closed path is zero. It is expressed as, ∫ E.dl = 0. The field having this property is called conservative or lamellar field.
Answer: d
See lessExplanation: Work done in moving a charge in a closed path is zero. It is expressed as, ∫ E.dl = 0. The field having this property is called conservative or lamellar field.
The potential of a uniformly charged line with density λ is given by, λ/(2πε) ln(b/a). State True/False.
Answer: a Explanation: The electric field intensity is given by, E = λ/(2πεr) Vab = -∫ E.dr = -∫ λ/(2πεr). On integrating from b to a, we get λ/(2πε) ln(b/a).
Answer: a
See lessExplanation: The electric field intensity is given by, E = λ/(2πεr)
Vab = -∫ E.dr = -∫ λ/(2πεr). On integrating from b to a, we get λ/(2πε) ln(b/a).
Find the potential between a(-7,2,1) and b(4,1,2). Given E = (-6y/x2 )i + ( 6/x) j + 5 k.
Answer: c Explanation: V = -∫ E.dl = -∫ (-6y/x2 )dx + ( 6/x)dy + 5 dz, from b to a. On integrating, we get -8.214 volts.
Answer: c
See lessExplanation: V = -∫ E.dl = -∫ (-6y/x2 )dx + ( 6/x)dy + 5 dz, from b to a.
On integrating, we get -8.214 volts.