Test: Line Integral - Electrical Engineering (EE) MCQ

Answer: b
Explanation: W = -Q E.dl
W = -2 X 10^-3 X (6y²z i + 12xyz j + 6xy² k) . (-3 i + 5 j -2 k)
At p(0,2,5), W = -2(-18.22.5) X 10^-3 = 0.72 J.

Answer: a
Explanation: Vab = -∫ E.dl is the relation between potential and field. It is clear that it is given by line integral.

The electric field (E) at a point in space is the force per unit charge experienced by a test charge at that point, hence the unit of E is newton per coulomb (N/C). The electric field is defined as E = - grad V

where grad is the gradient operator and V is the electric potential.

 If V = 2x²y - 5z

the electric field at point (-4, 3, 6) is given as follows:

E = - grad V

From the formula above, the gradient of V = (2x²y - 5z)

is given by grad V = (dV/dx)i + (dV/dy)j + (dV/dz)k

Where i, j and k are the unit vectors along the x, y and z axes respectively.

Thus,grad V = (4xy)i + (2x²)j - 5k

Now, evaluating at (-4, 3, 6)

we get the electric field,E = - grad V

=(-4)(3)(4)i + (2)(16)j - (5)k

= -48i + 32j - 5k

Therefore, the electric field at point (-4,3,6) is  E = -48i + 32j - 5k

which implies that the magniture of the electric field is given by sqrt[(-48)² + (32)² + (-5)²] = 57.905 N/C.

Answer: c
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x² dy + 2 dz) , from q to p.
On integrating, we get 106 volts.

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: 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: 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: b
Explanation: Work done in a lamellar field is zero. ∫ E.dl = 0,thus ∑V = 0. The potential will be zero.

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: 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.