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Page 1
For JEE Advanced
Objective Questions
More than One Correct Options
Q 1. Two concentric shells have radii R and 2R charges q
A
and q
B
and potentials 2Vand (3/2) V
respectively. Now shell B is earthed and let charges on them become q
A
' and q
B
'. Then
(a) q
A
/q
B
= 1/2 (b) q
A
'/q
B
'=1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
Q 2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms
-1
. There is a
uniform horizontal electric field of 10
4
N/C
(a) The horizontal range of the particle is 10 m (b) The time of flight of the particle is 2 s
(c) The maximum height reached is 5 m (d) The horizontal range of the particle is 5 m
Q 3. At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials
are 100 V and 75 V respectively. Then
(a) Potential at its surface is 150 V (b) The charge on the sphere is
10
50
10 C
3
?
?
(c) The electric field on the surface is 1500 V/m (d) The electric potential at its centre is 25 V
Q 4. Three charged particles are in equilibrium under their electrostatic forces only. Then
(a) The particles must be collinear (b) All the charges cannot have the same magnitude
(c) All the charges cannot have the same sign (d) The equilibrium is unstable
Q 5. Charges Q
1
and Q
2
lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and ?
be the flux of E over S
(a) If Q
1
changes, both E and ?
will change (b) If Q
2
changes, E will change but ?
will not change
(c) if Q
1
=0 and Q
2
? 0 then E ? 0 but ? = 0 (d) If Q
1
? 0and Q
2
= 0 then E = 0 but ? ? 0
Q 6. An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero
(b) The electric field is zero at every point of the sphere
(c) The electric field is not zero at any where on the sphere
(d) The electric field is zero on a circle on the sphere
Q 7. Mark the correct options.
(a) Gauss's law is valid only for uniform charge distributions
(b) Gauss's law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss's law is the field due to all the charges
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux
due to the charges enclosed by the surface
Page 2
For JEE Advanced
Objective Questions
More than One Correct Options
Q 1. Two concentric shells have radii R and 2R charges q
A
and q
B
and potentials 2Vand (3/2) V
respectively. Now shell B is earthed and let charges on them become q
A
' and q
B
'. Then
(a) q
A
/q
B
= 1/2 (b) q
A
'/q
B
'=1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
Q 2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms
-1
. There is a
uniform horizontal electric field of 10
4
N/C
(a) The horizontal range of the particle is 10 m (b) The time of flight of the particle is 2 s
(c) The maximum height reached is 5 m (d) The horizontal range of the particle is 5 m
Q 3. At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials
are 100 V and 75 V respectively. Then
(a) Potential at its surface is 150 V (b) The charge on the sphere is
10
50
10 C
3
?
?
(c) The electric field on the surface is 1500 V/m (d) The electric potential at its centre is 25 V
Q 4. Three charged particles are in equilibrium under their electrostatic forces only. Then
(a) The particles must be collinear (b) All the charges cannot have the same magnitude
(c) All the charges cannot have the same sign (d) The equilibrium is unstable
Q 5. Charges Q
1
and Q
2
lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and ?
be the flux of E over S
(a) If Q
1
changes, both E and ?
will change (b) If Q
2
changes, E will change but ?
will not change
(c) if Q
1
=0 and Q
2
? 0 then E ? 0 but ? = 0 (d) If Q
1
? 0and Q
2
= 0 then E = 0 but ? ? 0
Q 6. An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero
(b) The electric field is zero at every point of the sphere
(c) The electric field is not zero at any where on the sphere
(d) The electric field is zero on a circle on the sphere
Q 7. Mark the correct options.
(a) Gauss's law is valid only for uniform charge distributions
(b) Gauss's law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss's law is the field due to all the charges
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux
due to the charges enclosed by the surface
Q 8. Two concentric spherical shells have charges + q and - q as shown in figure. Choose the correct
options
(a) At A electric field is zero, but electric potential is non-zero
(b) At B electric field and electric potential both are non-zero
(c) At C electric field is zero but electric potential is non-zero
(d) At C electric field and electric potential both are zero
Q 9. A rod is hinged (free to rotate) at its centre O as shown in figure. Two point charges + q and + q
are kept at its two ends. Rod is placed in uniform electric field E as shown. Space is gravity free.
Choose the correct options.
(a) Net force from the hinge on the rod is zero
(b) Net force from the hinge on the rod is left wards
(c) Equilibrium of rod is neutral
(d) Equilibrium of rod is stable
Q 10. Two charges + Q each are fixed at points C and D. Line AB is the bisector line of CD. A third
charge + q is moved from A to B, then from B to C.
(a) From A to B electrostatic potential energy will decrease
(b) From A to B electrostatic potential energy will increase
(c) From B to C electrostatic potential energy will increase
(d) From B to C electrostatic potential energy will decrease
Answers
1.(a,d) 2.(a,b,c) 3.(a,b,c) 4.(all) 5.(a,b,c) 6.(a,c) 7.(c,d) 8.(a,b,d) 9.(b,c) 10.(b,c)
Solutions
1. (a) V
A
= 2V =
Solving these two equations we get
Page 3
For JEE Advanced
Objective Questions
More than One Correct Options
Q 1. Two concentric shells have radii R and 2R charges q
A
and q
B
and potentials 2Vand (3/2) V
respectively. Now shell B is earthed and let charges on them become q
A
' and q
B
'. Then
(a) q
A
/q
B
= 1/2 (b) q
A
'/q
B
'=1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
Q 2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms
-1
. There is a
uniform horizontal electric field of 10
4
N/C
(a) The horizontal range of the particle is 10 m (b) The time of flight of the particle is 2 s
(c) The maximum height reached is 5 m (d) The horizontal range of the particle is 5 m
Q 3. At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials
are 100 V and 75 V respectively. Then
(a) Potential at its surface is 150 V (b) The charge on the sphere is
10
50
10 C
3
?
?
(c) The electric field on the surface is 1500 V/m (d) The electric potential at its centre is 25 V
Q 4. Three charged particles are in equilibrium under their electrostatic forces only. Then
(a) The particles must be collinear (b) All the charges cannot have the same magnitude
(c) All the charges cannot have the same sign (d) The equilibrium is unstable
Q 5. Charges Q
1
and Q
2
lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and ?
be the flux of E over S
(a) If Q
1
changes, both E and ?
will change (b) If Q
2
changes, E will change but ?
will not change
(c) if Q
1
=0 and Q
2
? 0 then E ? 0 but ? = 0 (d) If Q
1
? 0and Q
2
= 0 then E = 0 but ? ? 0
Q 6. An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero
(b) The electric field is zero at every point of the sphere
(c) The electric field is not zero at any where on the sphere
(d) The electric field is zero on a circle on the sphere
Q 7. Mark the correct options.
(a) Gauss's law is valid only for uniform charge distributions
(b) Gauss's law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss's law is the field due to all the charges
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux
due to the charges enclosed by the surface
Q 8. Two concentric spherical shells have charges + q and - q as shown in figure. Choose the correct
options
(a) At A electric field is zero, but electric potential is non-zero
(b) At B electric field and electric potential both are non-zero
(c) At C electric field is zero but electric potential is non-zero
(d) At C electric field and electric potential both are zero
Q 9. A rod is hinged (free to rotate) at its centre O as shown in figure. Two point charges + q and + q
are kept at its two ends. Rod is placed in uniform electric field E as shown. Space is gravity free.
Choose the correct options.
(a) Net force from the hinge on the rod is zero
(b) Net force from the hinge on the rod is left wards
(c) Equilibrium of rod is neutral
(d) Equilibrium of rod is stable
Q 10. Two charges + Q each are fixed at points C and D. Line AB is the bisector line of CD. A third
charge + q is moved from A to B, then from B to C.
(a) From A to B electrostatic potential energy will decrease
(b) From A to B electrostatic potential energy will increase
(c) From B to C electrostatic potential energy will increase
(d) From B to C electrostatic potential energy will decrease
Answers
1.(a,d) 2.(a,b,c) 3.(a,b,c) 4.(all) 5.(a,b,c) 6.(a,c) 7.(c,d) 8.(a,b,d) 9.(b,c) 10.(b,c)
Solutions
1. (a) V
A
= 2V =
Solving these two equations we get
(b)
(c) & (d) Potential difference between A and B will remain unchanged as by earthing B, charge on
will not change.
2.
= 10m
3.
Solving these equations we get, and
R = 0.1 m
(a)
= 150V
(c)
= 1500 V/m
(d) V
centre
= 1.5 V
surface
4. See the hint of Q. No. 6 of subjective questions.
5. Electric field at any point depends on both charges Q
1
and Q
2
. But electric flux passing from any
closed surface depends on the charged enclosed by that closed surface only.
6. Flux from any closed surface due to a dipole.
7. Solution is not required.
8.
E
A
= E
C
= 0 but E
B
? 0
Page 4
For JEE Advanced
Objective Questions
More than One Correct Options
Q 1. Two concentric shells have radii R and 2R charges q
A
and q
B
and potentials 2Vand (3/2) V
respectively. Now shell B is earthed and let charges on them become q
A
' and q
B
'. Then
(a) q
A
/q
B
= 1/2 (b) q
A
'/q
B
'=1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
Q 2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms
-1
. There is a
uniform horizontal electric field of 10
4
N/C
(a) The horizontal range of the particle is 10 m (b) The time of flight of the particle is 2 s
(c) The maximum height reached is 5 m (d) The horizontal range of the particle is 5 m
Q 3. At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials
are 100 V and 75 V respectively. Then
(a) Potential at its surface is 150 V (b) The charge on the sphere is
10
50
10 C
3
?
?
(c) The electric field on the surface is 1500 V/m (d) The electric potential at its centre is 25 V
Q 4. Three charged particles are in equilibrium under their electrostatic forces only. Then
(a) The particles must be collinear (b) All the charges cannot have the same magnitude
(c) All the charges cannot have the same sign (d) The equilibrium is unstable
Q 5. Charges Q
1
and Q
2
lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and ?
be the flux of E over S
(a) If Q
1
changes, both E and ?
will change (b) If Q
2
changes, E will change but ?
will not change
(c) if Q
1
=0 and Q
2
? 0 then E ? 0 but ? = 0 (d) If Q
1
? 0and Q
2
= 0 then E = 0 but ? ? 0
Q 6. An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero
(b) The electric field is zero at every point of the sphere
(c) The electric field is not zero at any where on the sphere
(d) The electric field is zero on a circle on the sphere
Q 7. Mark the correct options.
(a) Gauss's law is valid only for uniform charge distributions
(b) Gauss's law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss's law is the field due to all the charges
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux
due to the charges enclosed by the surface
Q 8. Two concentric spherical shells have charges + q and - q as shown in figure. Choose the correct
options
(a) At A electric field is zero, but electric potential is non-zero
(b) At B electric field and electric potential both are non-zero
(c) At C electric field is zero but electric potential is non-zero
(d) At C electric field and electric potential both are zero
Q 9. A rod is hinged (free to rotate) at its centre O as shown in figure. Two point charges + q and + q
are kept at its two ends. Rod is placed in uniform electric field E as shown. Space is gravity free.
Choose the correct options.
(a) Net force from the hinge on the rod is zero
(b) Net force from the hinge on the rod is left wards
(c) Equilibrium of rod is neutral
(d) Equilibrium of rod is stable
Q 10. Two charges + Q each are fixed at points C and D. Line AB is the bisector line of CD. A third
charge + q is moved from A to B, then from B to C.
(a) From A to B electrostatic potential energy will decrease
(b) From A to B electrostatic potential energy will increase
(c) From B to C electrostatic potential energy will increase
(d) From B to C electrostatic potential energy will decrease
Answers
1.(a,d) 2.(a,b,c) 3.(a,b,c) 4.(all) 5.(a,b,c) 6.(a,c) 7.(c,d) 8.(a,b,d) 9.(b,c) 10.(b,c)
Solutions
1. (a) V
A
= 2V =
Solving these two equations we get
(b)
(c) & (d) Potential difference between A and B will remain unchanged as by earthing B, charge on
will not change.
2.
= 10m
3.
Solving these equations we get, and
R = 0.1 m
(a)
= 150V
(c)
= 1500 V/m
(d) V
centre
= 1.5 V
surface
4. See the hint of Q. No. 6 of subjective questions.
5. Electric field at any point depends on both charges Q
1
and Q
2
. But electric flux passing from any
closed surface depends on the charged enclosed by that closed surface only.
6. Flux from any closed surface due to a dipole.
7. Solution is not required.
8.
E
A
= E
C
= 0 but E
B
? 0
9.
Higher force = 2qE (toward left)
If we displace the rod, in displaced position too. Hence equilibrium is neutral.
10. Along the line AB, charge q is at unstable equilibrium position at B (When displaced from B
along AB, net force on it is away from B, whereas force at B is zero). Hence potential energy at B
is maximum.
Along CD equilibrium of q is stable. Hence potential energy at B is minimum along CD.
Match the Columns
Q 1. Five identical charges are kept at five vertices of a regular hexagon. Match the following two
columns at centre of the hexagon. If in the given situation electric field at centre is E. Then
Column I Column. II
(a) If charge at B is removed then electric field will become (P) 2E
(b) If charge at C is removed then electric field will become (q) E
(c) If charge at D is removed then electric field will become (r) zero
(d) If charges at B and C both are removed then electric field will become
(s) 3E
Note Only magnitudes of electric field are given.
Q 2. In an electric field
ˆˆ
E (2i 4j)N/ C, ??
?
electric potential at origin is 0 V. Match the following two
columns.
Column I Column II
Page 5
For JEE Advanced
Objective Questions
More than One Correct Options
Q 1. Two concentric shells have radii R and 2R charges q
A
and q
B
and potentials 2Vand (3/2) V
respectively. Now shell B is earthed and let charges on them become q
A
' and q
B
'. Then
(a) q
A
/q
B
= 1/2 (b) q
A
'/q
B
'=1
(c) potential of A after earthing becomes (3/2) V
(d) potential difference between A and B after earthing becomes V/2
Q 2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms
-1
. There is a
uniform horizontal electric field of 10
4
N/C
(a) The horizontal range of the particle is 10 m (b) The time of flight of the particle is 2 s
(c) The maximum height reached is 5 m (d) The horizontal range of the particle is 5 m
Q 3. At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials
are 100 V and 75 V respectively. Then
(a) Potential at its surface is 150 V (b) The charge on the sphere is
10
50
10 C
3
?
?
(c) The electric field on the surface is 1500 V/m (d) The electric potential at its centre is 25 V
Q 4. Three charged particles are in equilibrium under their electrostatic forces only. Then
(a) The particles must be collinear (b) All the charges cannot have the same magnitude
(c) All the charges cannot have the same sign (d) The equilibrium is unstable
Q 5. Charges Q
1
and Q
2
lie inside and outside respectively of a closed surface S. Let E be the field at
any point on S and ?
be the flux of E over S
(a) If Q
1
changes, both E and ?
will change (b) If Q
2
changes, E will change but ?
will not change
(c) if Q
1
=0 and Q
2
? 0 then E ? 0 but ? = 0 (d) If Q
1
? 0and Q
2
= 0 then E = 0 but ? ? 0
Q 6. An electric dipole is placed at the centre of a sphere. Mark the correct options.
(a) The flux of the electric field through the sphere is zero
(b) The electric field is zero at every point of the sphere
(c) The electric field is not zero at any where on the sphere
(d) The electric field is zero on a circle on the sphere
Q 7. Mark the correct options.
(a) Gauss's law is valid only for uniform charge distributions
(b) Gauss's law is valid only for charges placed in vacuum
(c) The electric field calculated by Gauss's law is the field due to all the charges
(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux
due to the charges enclosed by the surface
Q 8. Two concentric spherical shells have charges + q and - q as shown in figure. Choose the correct
options
(a) At A electric field is zero, but electric potential is non-zero
(b) At B electric field and electric potential both are non-zero
(c) At C electric field is zero but electric potential is non-zero
(d) At C electric field and electric potential both are zero
Q 9. A rod is hinged (free to rotate) at its centre O as shown in figure. Two point charges + q and + q
are kept at its two ends. Rod is placed in uniform electric field E as shown. Space is gravity free.
Choose the correct options.
(a) Net force from the hinge on the rod is zero
(b) Net force from the hinge on the rod is left wards
(c) Equilibrium of rod is neutral
(d) Equilibrium of rod is stable
Q 10. Two charges + Q each are fixed at points C and D. Line AB is the bisector line of CD. A third
charge + q is moved from A to B, then from B to C.
(a) From A to B electrostatic potential energy will decrease
(b) From A to B electrostatic potential energy will increase
(c) From B to C electrostatic potential energy will increase
(d) From B to C electrostatic potential energy will decrease
Answers
1.(a,d) 2.(a,b,c) 3.(a,b,c) 4.(all) 5.(a,b,c) 6.(a,c) 7.(c,d) 8.(a,b,d) 9.(b,c) 10.(b,c)
Solutions
1. (a) V
A
= 2V =
Solving these two equations we get
(b)
(c) & (d) Potential difference between A and B will remain unchanged as by earthing B, charge on
will not change.
2.
= 10m
3.
Solving these equations we get, and
R = 0.1 m
(a)
= 150V
(c)
= 1500 V/m
(d) V
centre
= 1.5 V
surface
4. See the hint of Q. No. 6 of subjective questions.
5. Electric field at any point depends on both charges Q
1
and Q
2
. But electric flux passing from any
closed surface depends on the charged enclosed by that closed surface only.
6. Flux from any closed surface due to a dipole.
7. Solution is not required.
8.
E
A
= E
C
= 0 but E
B
? 0
9.
Higher force = 2qE (toward left)
If we displace the rod, in displaced position too. Hence equilibrium is neutral.
10. Along the line AB, charge q is at unstable equilibrium position at B (When displaced from B
along AB, net force on it is away from B, whereas force at B is zero). Hence potential energy at B
is maximum.
Along CD equilibrium of q is stable. Hence potential energy at B is minimum along CD.
Match the Columns
Q 1. Five identical charges are kept at five vertices of a regular hexagon. Match the following two
columns at centre of the hexagon. If in the given situation electric field at centre is E. Then
Column I Column. II
(a) If charge at B is removed then electric field will become (P) 2E
(b) If charge at C is removed then electric field will become (q) E
(c) If charge at D is removed then electric field will become (r) zero
(d) If charges at B and C both are removed then electric field will become
(s) 3E
Note Only magnitudes of electric field are given.
Q 2. In an electric field
ˆˆ
E (2i 4j)N/ C, ??
?
electric potential at origin is 0 V. Match the following two
columns.
Column I Column II
(a) Potential at (4m. 0) (p.) 8 V
(b) Potential at (-4m, 0) (q) -8V
(c) Potential at (0, 4m) (r) 16 V
(d) Potential at (0, -4 m) (s) -16 V
Q 3. Electric potential on the surface of a solid sphere of charge is V. Radius of the sphere is 1 m.
Match the following two columns.
Column I Column II
(a) Electric potential at
R
r
2
? (p)
V
4
(b) Electric potential at r = 2R
(q)
V
2
(c) Electric field at
R
r
2
? (r)
3V
4
(d) Electric field at r = 2R (s) None of these
Q 4. Match the following two columns.
Column I Column II
(a) Electric potential (p) [MLT
-3
A
-1
]
(b) Electrical field (q) [ML
3
T
-3
A
-1
]
(c) Electric flux (r) [ML
2
T
-3
A
-1
]
(d) Permittivity of free space (s) None of these
Q 5. Match the following two columns.
Column I Column II
(a) Electric field due to charged spherical shell (P)
(b) Electric potential due to charged spherical shell (q)
(c) Electric field due to charged solid sphere (r)
(d) Electric potential due to charged solid sphere (s) None of these
Answers
1. (a) ? s (b) ? q (c) ? r (d) ? p 2. (a) ? q (b) ? p (c) ? s (d) ? r
3. (a) ? s (b) ? q (c) ? q (d) ? p 4. (a) ? r (b) ? p (c) ? q (d) ? s
5. (a) ? p (b) ? q (c) ? r (d) ? s
Solutions
1, (a) are cancelled. at 60°
(b) are cancelled. at 120°.
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FAQs on DC Pandey Solutions (JEE Advance): Electrostatic- 2 - DC Pandey Solutions for JEE Physics
| 1. What is the importance of studying electrostatics in the JEE Advanced exam? |  |
Ans. Studying electrostatics is crucial for the JEE Advanced exam as it forms the foundation for understanding various concepts in physics, especially in the field of electromagnetism. Questions related to Coulomb's law, electric fields, electric potential, and capacitance are frequently asked in the exam.
| 2. How can I effectively prepare for the electrostatics section of the JEE Advanced exam? | |
Ans. To effectively prepare for the electrostatics section, it is important to thoroughly understand the fundamental concepts and principles. Practice solving a variety of numerical problems and conceptual questions from reputable study materials like DC Pandey's book. Additionally, revising the formulas and regularly taking mock tests will help in strengthening your preparation.
| 3. Are there any specific problem-solving techniques that can be applied to solve electrostatics questions in the JEE Advanced exam? | |
Ans. Yes, there are certain problem-solving techniques that can be helpful in solving electrostatics questions. One such technique is the method of superposition, which involves breaking down complex systems into simpler ones and then adding their effects. Another technique is the use of Gauss's law to calculate the electric field in symmetric charge distributions. Additionally, understanding the concept of equipotential surfaces can simplify problem-solving.
| 4. Are there any common misconceptions or tricky concepts in the electrostatics section of the JEE Advanced exam? | |
Ans. Yes, there are a few common misconceptions and tricky concepts in the electrostatics section. One such misconception is incorrectly applying the principle of superposition without considering the direction of electric fields. Another tricky concept is understanding the behavior of conductors and insulators in the presence of electric fields. It is important to carefully analyze the given information and apply the correct principles to avoid falling into such traps.
| 5. How can the book "Electrostatics: JEE Advance(Part - 2) - Physics, Solution by DC Pandey JEE" help in preparing for the electrostatics section of the JEE Advanced exam? | |
Ans. The book "Electrostatics: JEE Advance(Part - 2) - Physics, Solution by DC Pandey JEE" is a valuable resource for JEE Advanced aspirants. It provides comprehensive explanations, solved examples, and practice questions specifically tailored to the exam syllabus. The book's content is designed to enhance conceptual understanding and problem-solving skills in electrostatics, making it an essential tool for effective preparation.
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Dec 26, 2024 Last updated |
Document Description: DC Pandey Solutions (JEE Advance): Electrostatic- 2 for JEE 2024 is part of DC Pandey Solutions for JEE Physics preparation.
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