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JEE Advanced Electrostatics Important Questions: Electrostatics is a crucial chapter in Physics for JEE Advanced. Electrostatics accounts for around 6% to 7% of the weightage in the JEE Advanced exam, and it is also extremely important in the final examination. When compared to the previous units, this is one of the simpler chapters.
Candidates can learn more about charge theory and properties, material classification by charge conductor, and electric potential due to the system of charges. This article contains all of the relevant electrostatics JEE Advanced questions. Physics is one of the most difficult subjects in PCM since it is based upon concepts. In an exam like JEE Advanced, questions are asked with a variety of ideas, making it difficult to determine which type of topic is being asked. Therefore, read this article to learn more about the JEE Advanced electrostatics questions.
The provisional answer keys of JEE Advanced 2022 will be released on September 3, 2022. The final JEE Advanced answer key will be released on September 11, 2022.
JEE Advanced exam 2022 was conducted on August 28, 2022.
Electromagnetism is a crucial component of the Class 12th syllabus and one of the most heavily weighted parts of JEE Advanced Physics. This entire module necessitates a great deal of practice and problem-solving abilities.
Electrostatics starts with Coulomb’s law and the principle of superposition before moving on to the concept of the electric field, electrostatic potential, and other principles that constitute the foundation of electromagnetic. These are the core principles, and students should fully comprehend them. All of these topics necessitate much practice. This topic contains simple questions in JEE Advanced, which can be readily cracked if one is conceptually strong.
The concept of conductors in electrostatics is a fundamental part of electrostatics. The majority of electrostatics problems are based on the characteristics of conductors under electrostatic conditions. As a result, it is essential that students practise a large number of questions from this area.
Another crucial concept is the concept of electrostatic flux and Gauss’ law. Both of these concepts are critical, and they can be linked to other topics. In JEE Advanced, there is always a question that involves calculating the flux of a charge via some or all surfaces.
Here are some electrostatics JEE Advanced questions for all JEE Advanced aspirants.
Q.1: Three charges +Q, q, +Q are placed, respectively, at a distance 0, d/2 and d from the origin on the x-axis. If the net force experienced by +Q placed at x = 0 is zero, then the value of q is
QQ/d2 + Qq/(d/2)2 =0
Q + 4q = 0
or q = -Q/4
Answer: (d) –Q/4
Q.2: A parallel plate capacitor having a capacitance of 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected, and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is
(a) 508 pJ
(b) 692 pJ
(c) 560 pJ
(d) 600 pJ
Initial Energy of the capacitor, Ui = (1/2) CV2
= (1/2) x 12pF x 10 x 10
= 600 pJ
After the slab, the energy of the slab, Uf = (1/2) Q2/C’
Q = CV = (12 pF)(10 V) = 120 p C
C’ = kC = 6.5 x 120 x 10-12 F
Therefore, Uf = [(1/2) (120 x 10-2)2]/[6.5 x 120 x 10-12]
Uf = 92 pJ
W + Uf= Ui
⇒ W = Ui – Uf
= 600 pJ – 92 pJ
= 508 pJ
Answer: (a) 508 pJ
Q.3: An electric field of 1000 V/m is applied to an electric dipole at an angle of 45°. The value of the electric dipole moment is 10–29 Cm. What is the potential energy of the electric dipole?
(a) –10 × 10–29 J
(b) –7 × 10–27 J
(c) –20 × 10–18 J
(d) –9 × 10–20 J
E = 1000 V/m , p = 10-29 cm, θ = 450
The potential energy stored in the dipole,
U = -p.Ecos θ = – 10-29 x 1000 x cos450
U = – 0.707 x 10-26 J= -7 x 10 -27 J
Answer: (b) –7 × 10–27 J
Q.4: The voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V m–1. The plate area is 10–4 m2. What is the dielectric constant if the capacitance is 15 pF? (given 0 = 8.86 × 10–12 C2 N–1 m–2)
C = Kε0A/d and V = Ed
Or K = CV/ε0AEmax
K = (15 x 10-12 x 500)/(8.86 x 10-12 x 10-4 x 106) = 8.5
Answer: (b) 8.5
Q.5: The bob of a simple pendulum has a mass of 2 g and a charge of 5.0 C. It is at rest in a uniform horizontal electric field of intensity 2000 V m–1. At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10 m s–2)
(a) tan–1 (0.2)
(b) tan–1 (0.5)
(c) tan–1 (2.0)
(d) tan–1 (5.0)
The forces acting on the bob are its weight and the force due to the field.
Tcosθ = mg ——-(1)
Tsinθ = qE ——–(2)
Dividing (2) by (1)
tanθ = qE/mg
θ = tan-1((5 x 10-6 x 2 x 103) / (2 x 10-3 x 10)) = tan-1(0.5)
Answer: (b) tan-1(0.5)
Q.6: A capacitor with a capacitance of 5 µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done?
(a) 6.25 × 10–6 J
(b) 3.75 × 10–6 J
(c) 2.16 × 10–6 J
(d) 2.55 × 10–6 J
Work done = Uf – Ui = (½)q2/Cf – (½)q2/Ci
Work done = q2/2[1/Cf – 1/Ci]
Work done = [(5 x 10-6)2/2][(1/(2 x 10-6)) – (1/(5 x 10-6))]
Work done = 3.75 x 10-6 J
Answer: (b) 3.75 × 10–6 J
Q.7: A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be
(a) 1.2 nC
(b) 0.3 nC
(c) 2.4 nC
(d) 0.9 nC
Induced charge on dielectric,
Qind = Q(1 – 1/K)
Final charge on the capacitor, Q = K C0V
Q = (5/3) x 90 x 10-12 x 20 = 3 x 10-9 C = 3nC
Qind = 3(1 – ⅗) = 3 x ⅖ = 1.2 nC
Answer: (a) 1.2 nC
Q.8: The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 μC charges, its radius will be [Take: (1/4πε0 ) = 9 x 109 Nm2/C2]
(a) 32 mm
(b) 20 mm
(c) 16 mm
(d) 28 mm
The energy stored in the electric field produced by a metal sphere = 4.5 J
⇒ Q2/2C = 4.5 or C = Q2/2 x 4.5
The capacitance of spherical conductor = 4πε0R
4πε0R = Q2/(2 x 4.5)
R = (1/4πε0) x [(4 x 10-6)2/(2 x 4.5)] = 9 x 109 x (16/9) x 10-12 = 16 x 10-3 m = 16 mm
Answer:(c) 16 mm
Q.9: There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P in the region is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field?
(a) 589.2 V
(b) 589.6 V
(c) 589.5 V
(d) 589.4 V
ΔV = E.d
ΔV = Edcosθ = 0.8 x cos 600
ΔV = 0.4
Hence the new potential at the point on the sphere is
589.0 + 0.4 = 589.4 V
Answer: (d) 589.4 V
Q.10: Two identical conducting spheres A and B, carry an equal charge. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to
Initially, force between spheres A and B, F = kq2/r
When A and C are touched, the charge on both will be q/2
Again C has touched with B the charge on B is given by
qB = ((q/2) + q)/2 = 3q/4
The required force between spheres A and B is given by
F’ = kqAqB/r2 = [k x (q/2) x (3q/4)]/r2 = (⅜)(kq2/r2) = ⅜F
Answer: (a) 3F/8
While electrostatics is an extensive subject covering a wide range of topics, we can divide it into the following two broad categories keeping the syllabus of JEE Advanced in mind. These are electrostatics Class 12 important questions that are also useful for JEE Advanced.
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Check below the frequently asked questions regarding electrostatics JEE Advanced questions:
Ans: Yes, the electrostatics chapter is important for JEE Advanced.
Ans: Electrostatics is said to have a weightage of 3.3% in JEE Advanced.
Ans: Yes, it is tough and requires a lot of practice and strong conceptual knowledge.
Ans: Yes, questions repeat, but the exam pattern does not change.
Ans: JEE Advanced 2022 was held on August 28, 2022.
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