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For a test charge placed at infinity, the electric field will be
Answer: c Explanation: E = Q/ (4∏εor2 ) When distance d is infinity, the electric field will be zero, E= 0.
Answer: c
Explanation: E = Q/ (4∏εor2
)
When distance d is infinity, the electric field will be zero, E= 0.
See lessThe lines of force are said to be
Answer: c Explanation: The lines drawn to trace the direction in which a positive test charge will experience force due to the main charge are called lines of force. They are not real but drawn for our interpretation.
Answer: c
Explanation: The lines drawn to trace the direction in which a positive test charge will
experience force due to the main charge are called lines of force. They are not real but
drawn for our interpretation.
See lessThe electric flux density is the
Answer: a Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric field intensity. Thus electric flux density is the product of permittivity and electric field intensity.
Answer: a
Explanation: D= εE, where ε=εoεr is the permittivity of electric field and E is the electric
field intensity. Thus electric flux density is the product of permittivity and electric field
intensity.
See lessWhich of the following correctly states Gauss law?
Answer: d Explanation: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density.
Answer: d
Explanation: The electric flux passing through any closed surface is equal to the total
charge enclosed by that surface. In other words, electric flux per unit volume leaving a
point (vanishing small volume), is equal to the volume charge density.
See lessThe Gaussian surface is
Answer: b Explanation: It is any physical or imaginary closed surface around a charge which satisfies the following condition: D is everywhere either normal or tangential to the surface so that D.ds becomes either Dds or 0 respectively.
Answer: b
Explanation: It is any physical or imaginary closed surface around a charge which
satisfies the following condition: D is everywhere either normal or tangential to the
surface so that D.ds becomes either Dds or 0 respectively.
See lessFor a charge Q1, the effect of charge Q2 on Q1 will be,
Answer: b Explanation: The force of two charges with respect with each other is given by F1 and F2. Thus F1 + F2 = 0 and F1 = -F2.
Answer: b
Explanation: The force of two charges with respect with each other is given by F1 and
F2. Thus F1 + F2 = 0 and F1 = -F2.
See lessThe electric field intensity is defined as
Answer: c Explanation: The electric field intensity is the force per unit charge on a test charge, i.e, q1 = 1C. E = F/Q = Q/(4∏εr2).
Answer: c
Explanation: The electric field intensity is the force per unit charge on a test charge, i.e,
q1 = 1C. E = F/Q = Q/(4∏εr2).
See lessFind the force on a charge 2C in a field 1V/m.
Answer: c Explanation: Force is the product of charge and electric field. F = q X E = 2 X 1 = 2 N.
Answer: c
Explanation: Force is the product of charge and electric field.
F = q X E = 2 X 1 = 2 N.
See lessFind the electric field intensity of two charges 2C and -1C separated by a distance 1m in air.
Answer: b Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109 E = F/q = 18 X 109 /2 = 9 X 109 .
Answer: b
Explanation: F = q1q2/(4∏εor2) = -2 X 9/(10-9 X 12) = -18 X 109
E = F/q = 18 X 109
/2 = 9 X 109
.
See lessThe divergence theorem value for the function x2 + y2 + z2 at a distance of one unit from the origin is
Answer: d Explanation: Div (F) = 2x + 2y + 2z. The triple integral of the divergence of the function is ∫∫∫(2x + 2y + 2z)dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.
Answer: d
Explanation: Div (F) = 2x + 2y + 2z. The triple integral of the divergence of the function is
∫∫∫(2x + 2y + 2z)dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3
units.
See less