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# JEE Advanced 2023: Questions on Electrostatics

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.

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

(a) +Q/4

(b) –Q/2

(c) +Q/2

(d) –Q/4

Solution

QQ/d2 + Qq/(d/2)2 =0

Q + 4q = 0

or q = -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

Solution

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

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

Solution

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)

(a) 3.8

(b) 8.5

(c) 6.2

(d) 4.5

Solution

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

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)

Solution

The forces acting on the bob are its weight and the force due to the field.

At equilibrium,

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)

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

Solution

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

Solution

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

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

Solution

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

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

Solution

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

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

(a) 3F/8

(b) F/2

(c) 3F/4

(d) F

Solution

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

### Important Topics in Electrostatics Class 12

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.

#### Electric Charges and Field

• Definition of the electric dipole with SI and formula
• Gauss law and its applications with some numerical from NCERT and SL ARORA.
• Electric field due to a dipole at its axial and equatorial position with their ratio and proper diagram
• Some numerical based on the above topic and must do that type of question which ask where the electric field/ potential is zero. Also, do triangle and square type questions.
• Torque on doppler, SI unit, numerical.

#### Electric Potential and Capacitance

• Definition of electronic potential, formula, SI unit, quantity
• Work done in rotating dipole (learn special cases with numerical)
• Combination of capacitors with dielectrics constant K
• Numericals on capacitors are based on shortness, and symmetry.
• Charge in capacitors and dedication of force between the plates of capacitors.

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### FAQs Regarding JEE Advanced 2023 Electrostatics Important Questions

Q.1: Is electrostatics important for JEE Advanced?

Ans: Yes, the electrostatics chapter is important for JEE Advanced.

Q.2: How important is electrostatics in JEE Advanced?

Ans: Electrostatics is said to have a weightage of 3.3% in JEE Advanced.

Q.3: Is electrostatics tough for JEE Advanced?

Ans: Yes, it is tough and requires a lot of practice and strong conceptual knowledge.

Q.4: Do questions repeat from Electrostatics in JEE Advanced?

Ans: Yes, questions repeat, but the exam pattern does not change.

Q.5: When was the JEE Advanced 2022 exam held?

Ans: JEE Advanced 2022 was held on August 28, 2022.

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