Bihar BoardEnglish MediumSTD 12 SciencePhysicsELECTROSTATIC POTENTIAL AND CAPACITANCE2 Marks
Question
Explain (or write a short note on) capacitors and capacitance.
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Answer
→ A system of two conductors separated by a dielectric medium, is called a capacitor. → As shown in fig., two isolated/ or separated) conductors have charges $+Q$ and $-Q$ and their potentials are $V_1$ and $V_2$ respectively, and p.d. between them is $V=V_1-V_2$. → These conductors can be charged by connecting them to a battery. Charge on any one of the two conductors is called the charge of capacitor. → Total (Net) charge of the capacitor is zero. → The electric field in the region between the conductors is proportional to charge Q . ( $E \propto Q$ ) →The p.d. between two conductors means the work done per unit positive charge in taking a small test charge from the conductor 2 to conductor 1 against electric field. As the charge increases, potential also increases. → Consequently, $V \propto Q$ and therefore, the ratio $\frac{ Q }{ V }$ is constant. $\therefore C =\frac{ Q }{ V }$ where constant C is called the capacitance of the capacitor. → C is independent of Q and V . → C depends on geometric configuration (shape, size, separation) of the system of two conductors, and on the dielectric constant of the material placed between two conductors. → SI unit of capacitance is F (farad) OR C/V $1 F=1 C$. $V ^1$ → Its dimensional formula is $M ^{-1} L^{-2} T^4 A^2$ → Smaller units used in day to day life : $\begin{array}{l} 1 \mu F =10^{-6} F \quad 1 p F=10^{-12} F \\ 1 n F=10^{-9} F \end{array}$ → The circuit symbol of a capacitor with fixed capacitance is :
→ While the capacitor with variable capacitance is shown as
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