Fundamentals of Physics Chapter 25 Solutions: Capacitance

Halliday Resnick and Walker Fundamentals of Physics Volume 2 Solutions for Chapter 25 ‘Capacitance’ will help you prepare for Class 12 board exams as well as competitive exams like JEE. A major portion of the chapter covers the topic like determining capacitance, a combination of capacitance, capacitance in parallel and in series, energy stored in an electric field, a capacitor with a dielectric and Gauss’s Law.  Once you get thorough with the chapter you would be able to figure out a circuit with a parallel plate capacitor, calculate capacitance, the equivalent of parallel capacitors and equivalents of series capacitors.

Resnick Halliday & Walker Fundamentals of Physics Volume 2 Solutions for Capacitance consist of one exercise with 55 questions which are divided into six different modules of the chapter. The questions are about calculating capacitance, potential difference, finding equivalent capacitance, finding the amount of stored charge, calculating energy density, construction of capacitor. Moreover, you would be able to explain the relationship between potential energy, potential difference and a capacitance. Distinguishing polar and non-polar dielectrics and applying Gauss’ Law is the add on.

Our subject matter experts have furnished the excellent solutions for Resnick Halliday & Walker Volume 2 Capacitance for Class 12 with the chief motive to help you understand the concept easily. To achieve fantastic results in JEE and NEET, learning this chapter and going through all the problems is crucial. Our specialists have worked very hard to solve each problem without any complexity. This will help you take note of all the exceptions and key points of different topics in the chapter.

Important Topics for Halliday Resnick and Walker Fundamentals of Physics Volume 2 Solutions Chapter 25: Capacitance

What is physics?

Physics deals with basic science which will help to provide practical devices. This chapter focuses on the capacitor, a device which is used to store energy. The initial step in the discussion of capacitors is to find how much charge can be stored in it. This amount is called capacitance.


When two conductors are isolated electrically from each other and the surrounding, it forms a capacitor. A charged capacitor has the same magnitude q but with opposite signs. Despite being any size, shape or geometry, a conductor is called plates.

Two parallel plates capacitor separated by distance d. The charges on both the plates with the facing sides have magnitude with opposite signs. The magnitudes are +q and -q. Here, q is the absolute value of charges on plates. All points on plates have the same electric potential but there is a potential difference between two plates. Therefore, the charge q and the potential difference V are proportional to each other. So that becomes q=CV. 

C is a Capacitance which means how much a plate should be charged to achieve a certain potential difference. The greater the capacitance, the more charge it requires. 

Charging a Capacitor

Capacitors can be charged by placing it in an electric circuit with a battery. A device maintaining a certain potential difference between terminals is battery and A path through which charge flows is an electric circuit. A capacitor is said to be charged when the electric field is zero and there is no more drive of electrons.


  • Calculating Capacitance


Steps to calculate Capacitance are

  1. Assume charge q.
  2. Using Gauss’ Law, calculate the electric field between the plates.
  3. Calculate the potential difference V.
  4. Calculate Capacitance C.


  • Calculating Electric Field


q = ε0 E A

where A is an area of Gaussian flux surface and E will have uniform magnitude.


  • Calculating the potential difference


Where E and ds are vectors having opposite paths. Here, our path of integration starts on a negative plate and ends on a positive plate.


  • Determining Capacitance


When the capacitor has flat parallel plates and a certain distance between them, C = ε0A/d

For Cylindrical Capacitor of length L and radii a and b, capacitance:

C = 2πε0 L/log(b/a)

For spherical capacitor with radii a and b, capacitance:

For an isolated capacitor of radius R, capacitance:

C= 4πε0R