NCERT Solutions for Class 12 Physics Chapter 2
NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance Exercises and Additional Exercises in PDF format free download updated for new academic session 2020-2021 based on new NCERT Books.Download UP Board Solutions, NCERT Solutions and NCERT Apps based on updated CBSE Syllabus 2020-21. If you have any doubt, please visit to discussion forum to ask your doubts.
NCERT Solutions for Class 12 Physics Chapter 2
|Chapter 2:||Electrostatic Potential and Capacitance|
Chapter 2 Electrostatic Potential and Capacitance Solutions
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Class 12 Physics Chapter 2 Solutions in English
NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance exercises are given below to use it online or download in PDF form for offline. Ask your doubts related to NIOS or CBSE Board or any other educational fact through Discussion Forum and reply to the other users. Download NCERT Books 2020-21 based on latest CBSE Syllabus for all boards who are following CBSE.
Questions from Board Papers
1. What is the ratio of electric field intensity at a point on the equatorial line to the field at a point on axial line when the points are at the same distance from the centre of the dipole?
2. For an isolated parallel plate capacitor of capacitance C and potential difference V, what will happen to (i) charge on the plates (ii) potential difference across the plates (iii) field between the plates (iv) energy stored in the capacitor, when the distance between the plates is increased? [Answer: (i) No change (ii) increases (iii) No change (iv) increases].
3. A storage capacitor on a RAM (Random Access Memory) chip has a capacity of 55pF. If the capacitor is charged to 5.3V, how may excess electrons are on its negative plate?[Answer: 1.8 × 10^9]
4. The potential at a point A is –500V and that at another point B is +500V. What is the work done by external agent to take 2 units (S.I.) of negative charge from B to A.
5. In charging a capacitor of capacitance C by a source of emf V, energy supplied by the sources QV and the energy stored in the capacitor is ½QV. Justify the difference.
Important Questions for practice
1. A point charge Q is kept at the intersection of (i) face diagonals (ii) diagonals of a cube of side a. What is the electric flux linked with the cube in (i) & (ii)?
2. Using Gauss’s theorem in electrostatics, deduce an expression for electric field intensity due to a charged spherical shell at a point (i) inside (ii) on its surface (iii) outside it. Graphically show the variation of electric field intensity with distance from the centre of shell.
3. A conducting slab of thickness ‘t’ is introduced between the plates of a parallel plate capacitor, separated by a distance d (t is less than d). Derive an expression for the capacitance of the capacitor. What will be its capacitance when t is equal to d?
4. A potential difference V is applied across a conductor of length L and diameter D. How are the electric field E and the resistance R of the conductor affected when (i) V is halved (ii) L is halved (iii) D is doubled.
Justify your answer.
5. What is an equipotential surface? Write three properties Sketch equipotential surfaces of (i) Isolated point charge (ii) Uniform electric field (iii) Dipole
Important Questions on 12th Physics Chapter 2
Charge is uniformly distributed over the conductor, q = 1.6 × 10^−7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF
Therefore, total capacitance of the combination is 9 pF.
If voltage supply remained connected, voltage between two plates will be constant.
Supply voltage, V = 100 V
Initial capacitance, C = 1.771 × 10−11 F
New capacitance, C1 = kC = 6 × 1.771 × 10−11 F = 106 pF
New charge, q1 = C1V = 106 × 100 pC = 1.06 × 10–8 C
Potential across the plates remains 100 V.
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.
Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.