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If potential V = 20/(x2 + y2 ). The electric field intensity for V is 40(x i + y j)/(x2 + y2 ) 2 . State True/False.
Answer: a Explanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) =40(x i + y j)/(x2 + y2)2. Thus the statement is true.
Answer: a
See lessExplanation: E = -Grad (V) = -Grad(20/(x2 + y2)) = -(-40x i /(x2 + y2)2 – 40(y j)/(x2 + y2)2) =40(x i + y j)/(x2 + y2)2. Thus the statement is true.
Find the electric flux density surrounding a material with field intensity of 2xyz placed in transformer oil ( εr = 2.2) at the point P(1,2,3) is (in 10-10 units)
Answer: c Explanation: D = εE, where ε = εo εr. The flux density is given by, D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.
Answer: c
See lessExplanation: D = εE, where ε = εo εr. The flux density is given by,
D = 8.854 X 10-12 X 2.2 X 2(1)(2)(3) = 2.33 X 10-10 units.
Find the potential at a point (4, 3, -6) for the function V = 2x2y + 5z.
Answer: b Explanation: The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.
Answer: b
See lessExplanation: The electric potential for the function V = 2x2y + 5z at the point (4, 3, -6) is given by V = 2(4)2(3) + 5(-6) = 96-30 = 66 units.
Find the electric potential for an electric field 3units at a distance of 2m.
Answer: c Explanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.
Answer: c
See lessExplanation: The electric field intensity is the ratio of electric potential to the distance. E = V/d. To get V = E X d = 3 X 2 = 6units.
The electric field intensity is the negative gradient of the electric potential. State True/False.
Answer: a Explanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = - Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.
Answer: a
See lessExplanation: V = -∫E.dl is the integral form. On differentiating both sides, we get E = –
Grad (V). Thus the electric field intensity is the negative gradient of the electric potential.
The electric flux density and electric field intensity have which of the following relation?
Answer: a Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship
Answer: a Explanation: The electric flux density is directly proportional to electric field intensity. The proportionality constant is permittivity. D=ε E. It is clear that both are in linear relationship
See lessWith Gauss law as reference which of the following law can be derived?
Answer: c Explanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity. We know that F = qE. Hence force can be computed. This gives the Coulomb’s law.
Answer: c
See lessExplanation: From Gauss law, we can compute the electric flux density. This in turn can be used to find electric field intensity. We know that F = qE. Hence force can be
computed. This gives the Coulomb’s law.
The normal component of the electric flux density is always discontinuous at the interface. State True/False.
Answer: a Explanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.
Answer: a
See lessExplanation: In a dielectric-dielectric boundary, if a free surface charge density exists at the interface, then the normal components of the electric flux density are discontinuous at the boundary, which means Dn1 = Dn2.
The tangential component of electric field intensity is always continuous at the interface. State True/False.
Answer: a Explanation: Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.
Answer: a
See lessExplanation: Consider a dielectric-dielectric boundary, the electric field intensity in both the surfaces will be Et1 = Et2, which implies that the tangential component of electric field intensity is always continuous at the boundary.
Gauss law cannot be expressed in which of the following forms?
Answer: d Explanation: Gauss law can be expressed in differential or point form as, Div (D)= ρv and in integral form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.
Answer: d
See lessExplanation: Gauss law can be expressed in differential or point form as,
Div (D)= ρv and in integral form as ∫∫ D.ds = Q = ψ . It is not possible to express it using Stoke’s theorem.