Question

Chapter 3 Capacitors Learning Objectives By working through this lab, you will . examine how capacitance depends on parameters such as area, and dis- tance between plates for a parallel plate capacitor, examine how dielectric materials affect the capacitance of a capacitor, determine the dielectric constant of unknown dielectric materials. 3.1 Pre-lab In the lab on electric potential and electric field lines, you noted that charged par ticles produce electric fields and these electric fields, in turn, act on charged par- ticles. However, you then proceeded to produce an electric field geometry using conductors held at a particular electric potential. What happened to the electric charges? Electric charge and electric potential are closely related to each other. Putting charge on a conductor raises its electric potential. It is usually much easier to measure potential than to measure charge, so physicists often work with electric potential. However, charge and potential are proportional to each other. Consider the configuration given in Fig. 3.1. Figure 3.1- Charged Parallel Plates

Answer 1

Electric field lines are oriented from +ve plate to -ve plate as shown in figure let sigma = Q/A where A = surface area then electric field strength between plates E = sigma/2 eplisonvariant_0 - (- sigma/2 eplisonvariant_0) = sigma/ eplisonvariant_0 Potential difference between plate V = Ed = sigma d/e_0 put value of sigma V = Qd/A eplisonvariant_0 and we know that C = Q/v = Q/Qd eplisonvariant_0 A = eplisonvariant_0 A/D C = eplisonvariant_0 A/d \r\n In case I cottage decreases Electric field decreases Charge decreases Calogeitance remain constant In case II Voltage increases Electric field increases Charge increases Calracitance remain same In case II voltage, Electricfield, charge and capacitance all remain same dipole will try to align itself along electric field as shown New capacitance will be C_1 = k eplisonvariant_0A/d = ke where k = dielectric constant