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Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) PDF Download

Q1: As shown in the figure below, two concentric conducting spherical shells, centered at r = 0 and having radii r = c and r = d are maintained at potentials such that the potential V(r) at r = c is V1 and V(r) at r = d is V2. Assume that V(r) depends only on r, where r is the radial distance. The expression for V(r) in the region between r = c and r = d is     (2022)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

(b) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
(c) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
(d) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Ans: (b)
Sol: We have, Laplace equation
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)V will have only radial component.
From equation (1)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Integrate both side Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Again integrate,Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Here, P and Q are constant.
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)From eq. (2) & (3)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)From eq. (1),
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q2: Consider a large parallel plate capacitor. The gap 'd' between the two plates is filled entirely with a dielectric slab of relative permittivity 5. The plates are initially charged to a potential difference of V volts and then disconnected from the source. If the dielectric slab is pulled out completely, then the ratio of the new electric field E2 in the gap to the original electric field E1 is ________.        (2021)
(a)  2
(b) 5
(c) 6
(d) 8
Ans:
(b)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)If voltage source is removed then in both cases charge O is constant.
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Equation (i) is equal to equation (ii)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)⇒ 5V1 = V2
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Put equation (iii) in equation (iv),
⇒ E2/E1 = 5

Q3: A  1μC point charge is held at the origin of a cartesian coordinate system. If a second point charge o 10μC is moved from (0, 10, 0) to  (5, 5, 5) and subsequently to (5, 0, 0), then the total work done is ___________ mJ. (Round off to 2 decimal places).
Take Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) in SI units. All coordinates are in meters.      (2021)
(a) 5.25
(b) 9
(c) 8.62
(d) 2.47
Ans:
(b)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)In this case work done is independent of type of path but depends an intial and final point.
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Work done (by external source) in moving Q-charge (10 μC) in the presence of electric fieldPrevious Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q4: Let ar, aϕ and az be unit vectors along r, ϕ and z directions, respectively in the cylindrical coordinate system. For the electric flux density given by D = (ar15 + aϕ2r − az3rz) Coulomb/ m2, the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is:       (2020)
(a) 108 π
(b) 54 π
(c) 90 π
(d) 180 π
Ans
: (d)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q5: The static electric field inside a dielectric medium with relative permittivity,  ε= 2.25, expressed in cylindrical coordinate system is given by the following expression
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)where ar, aφ, az are unit vectors along r, φ and z directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, ε0, in SI units is given by:     (2020)
(a) 3ε0
(b) 4ε0
(c) 5ε0 
(d) 9ε0
Ans: (d)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
volume charge density
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q6: A co-axial cylindrical capacitor shown in Figure (i) has dielectric with relative permittivity εr1 = 2. When one-fourth portion of the dielectric is replaced with another dielectric ofrelative permittivity εr2, as shown in Figure (ii), the capacitance is doubled. The value of  εr12 is ____.     (2019)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) 10
(b) 7
(c) 15
(d) 18
Ans:
(a)
Sol: Co-axial cylindrical capacitor-1
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Co-axial cylindrical capacitor-2
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Put equation (i), (ii) in equation (iii),
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q7: The capacitance of an air-filled parallel-plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates, is placed on one of the plates covering it entirely, the capacitance becomes 86 pF.Neglecting the fringing effects, the relative permittivity of the dielectric is ______ (up to 2 decimal places).      (2018)
(a) 1.55
(b) 2.53
(c) 3.75
(d) 4.25
Ans:
(b)
Sol: Given, C = Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
In second case:
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Capacitance,
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q8: A positive charge of 1 nC is placed at (0, 0, 0.2) where all dimensions are in metres. Consider the x-y plane to be a conducting ground plane. Take ϵ0 = 8.85 × 10−12 F/m. The z component of the E field at (0, 0, 0.1) is closest to    (2018)
(a) 899.18 V/m
(b) −899.18 V/m
(c) 999.09 V/m
(d) −999.09 V/m
Ans:
(d)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Net electric field at point P due to charge Q is,
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q9: A thin soap bubble of radius R = 1 cm, and thickness a = 3.3μm(a < < R), is at a potential of 1 V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap in the thin bubble is 4πR2a and that of the drop is 4/3 πr3. The potential in volts, of the resulting single spherical drop with respect to the same reference point at infinity is __________. (Give the answer up to two decimal places.)      (SET-2 (2017))
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) 10.03
(b) 20.25
(c) 30.28
(d) 40.08
Ans:
(a)
Sol: After burst,
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Radius of soap drop = (3R2a)1/3 = 0.099665cm
Initial voltage was 1 V and C = 4πε0R
and initial charge, Q = Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Since after bursting Q remainsame and C = 4πε0r
New potential on soap drop,
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q10: Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed.     (SET-2  (2017))
(a) uniformly over the entire volume of the sphere
(b) uniformly over the outer surface of the sphere
(c) concentrated around the centre of the sphere
(d) along a straight line passing through the centre of the sphere
Ans: 
(b)
Sol: Added charge (one million electrons) to be solid spherical conductor is uniformly distributed over the outer surface of the sphere.

Q11: Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t = 0, the particles are released but are constrained to move along the same straight line. Which of these will collide first?     (SET-1 (2017))
(a) The particles will never collide
(b) All will collide together
(c) Proton and Neutron
(d) Electron and Neutron
Ans:
(d)
Sol: Given that electron, neutron and proton are in straight line.
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)The electron will move towards proton and proton will move towards electron and force will be same Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) But acceleration of electron will be more than proton as mass of electron < mass of proton. Since neutron are neutral they will not move. Thus electron will hit neutron first.

Q12: Two electrodes, whose cross-sectional view is shown in the figure below, are at the same potential. The maximum electric field will be at the point      (SET-2 (2016))
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) (A)
(b) (B)
(c) (C)
(d) (D)
Ans:
(a)
Sol: Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)At the point C, Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) (electric field intensity) is maximum being closest to the other plate.

Q13: A parallel plate capacitor filled with two dielectrics is shown in the figure below. If the electric field in the region A is 4 kV/cm, the electric field in the region B, in kV/cm, is     (SET-2 (2016))
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) 1
(b) 2
(c) 4
(d) 16
Ans:
(c)
Sol: As voltage is same across both the regions and distance between two plates is also same, then electric field remains same throughout the both region. As we know, E = V/d = constant for both regions.

Q14: Two electric charges q and -2q are placed at (0, 0) and (6, 0) on the x-y plane. The equation of the zero equipotential curve in the x - y plane is     (SET-1 (2016))
(a) x = -2
(b) y = 2
(c) x2 + y2 = 2
(d) (x + 2)+ y= 16
Ans: 
(d)
Sol: Charge, Q is located at (0, 0) and  −2θ is located at (6, 0).
To find V at any point (x, y)
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Q15: Two semi-infinite conducting sheets are placed at right angles to each other as shown in the figure. A point charge of +Q is placed at a distance of d from both sheets. The net force on the charge is Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) where K is given by      (SET-2 (2015))
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)(a) 0
(b) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

(c) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
(d) Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)
Ans: (d)
Sol: Due to image of the charge, three new charges will be created as shown in figure
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)Total force (4) due to (1), (2), (3) will be
Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

The document Previous Year Questions- Electrostatic Fields - 1 | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Electromagnetic Fields Theory (EMFT).
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FAQs on Previous Year Questions- Electrostatic Fields - 1 - Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

1. What is the concept of electric field in electrostatic fields?
Ans. The electric field is a physical quantity that describes the force experienced by a charged particle at a given point in space. It is a vector field that points in the direction of the force that a positive test charge would experience if placed at that point.
2. How is the electric field intensity calculated in electrostatic fields?
Ans. The electric field intensity at a point in space is calculated by dividing the force experienced by a positive test charge placed at that point by the magnitude of the test charge. Mathematically, it is represented as E = F/q, where E is the electric field intensity, F is the force, and q is the test charge.
3. What is Gauss's Law and how is it applied in electrostatic fields?
Ans. Gauss's Law states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface. It is a fundamental principle in electrostatics and is used to calculate electric fields in symmetric situations, such as spherical or cylindrical geometries.
4. What is the significance of electric potential in electrostatic fields?
Ans. Electric potential is a scalar quantity that represents the work done in bringing a unit positive charge from infinity to a point in space. It is used to define the concept of potential energy in an electric field and is essential in understanding the behavior of charged particles in electrostatic fields.
5. How does the presence of dielectric materials affect electric fields in electrostatics?
Ans. Dielectric materials have the ability to polarize in the presence of an electric field, which leads to the weakening of the field within the material. This results in the redistribution of charges and a decrease in the overall electric field intensity, affecting the behavior of the electric field in the presence of dielectrics.
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