Physics Vidyapith ✍️

Description→ When a capacitor is charged, the work is done by charged battery (i.e the chemical energy of the battery is used to charge the capacitor). As the capacitor gets charged, the potential difference between its plates increases. Due to this increase in the potential difference between the plates, the battery has to give the same amount of charge to the capacitor. Because of that, the battery has to do more and more work. The total amount of work done in charging the capacitor is stored in the form of electric potential energy in between the capacitor plates. This energy is retrieved as heat when the capacitor is discharged through a resistance. Derivation→ Let us consider, a capacitor of capacitance $C$with a potential difference of $V$ between the plates. In the process of charging, electrons are transferred from the positive to negative, unit each plate acquires an amount of charge $q$. Suppose during the process of charging, we increase the charge fr

Derivation and Description→ Consider, A parallel plate capacitor with a charge $+q$ on one of its plates and $-q$ on the other plate. Let initially the plates of the capacitor are almost, but not quite touching. Due to opposite polarity, there is an attractive force $F$ between the plates. Now If these plates are gradually pulling and apart to a distance $d$, in such a way that $d$ is still small compared to the linear dimension of the plates, then the approximation of a uniform field between the plates is maintained and thus the force remains constant. The force between the charged parallel plates of the capacitor Now the work done in separating the plates from near $0$ to $d$, $W=F.d \qquad(1)$ This work done $(W)$ is stored as electrostatic potential energy $(U)$ between the plates, i.e. $U=\frac{1}{2}q V $ But $V=E.d$, then the above equation can be written as $U=\frac{1}{2}q \: E \: d \qquad(2)$ But the equation $(1)$ and equation $(2)$ both

Derivation→ Let us consider, The charge on a parallel-plate capacitor = $q$ The area of parallel-plate = $A$ The distance between the parallel-plate = $d$ The dielectric constant of the slab of a material =$K$ The thickness of the material =$t$ The vacuum (or air) between the plates =$(d-t)$ The surface charge density on the plates= $\sigma$ The capacitance of Parallel Plate Capacitor with Dielectric Slab between Plates The electric field in the air between the plates is $E_{\circ}=\frac{\sigma}{\epsilon_{\circ}}$ $E_{\circ}=\frac{q}{\epsilon_{\circ}\: A} \qquad(1)$ The electric field in the dielectric material $E=\frac{q}{\epsilon_{\circ}\: K\: A} \qquad(2)$ The potential difference between the plates $V=E_{\circ}\left( d-t \right)+E\:t \qquad(3)$ Now substitute the value of $E_{\circ}$ and $E$ in the equation $(3)$, Then we get $V=\frac{q}{\epsilon_{\circ}\: A} \left( d-t \right)+ \frac{q}{\epsilon_{\c

Parallel Plate Air Capacitor→ A parallel-plate capacitor consists of two long, plane, metallic plates mounted on two insulating stands and placed at a small distance apart in a vacuum (or air). The plates are exactly parallel to each other. Parallel Plate Air Capacitor Derivation of the capacitance of parallel plate capacitor in the air→ Let us consider, Two plates $X$ and $Y$ are separated at a small distance $d$ in the vacuum (or air). If the area of plates is $A$ and the plates $X$ and $Y$ have charge $+q$ and $-q$ respectively. If the surface charge density on each plate is $\sigma$ then electric field intensity at a point between two parallel plates is $E=\frac{\sigma}{\epsilon_{\circ}}$ $E=\frac{q}{\epsilon_{\circ}A}\qquad(1) \qquad (\because \sigma=\frac{q}{A}) $ The potential difference between the parallel plates capacitor in air (or vacuum) is $V=E.d$ Now substitute

Definition : The work done in charging the conductor is stored as potential energy in the electric field in the vicinity of the conductor is called the potential energy of a charged conductor. Derivation → Let us consider, a conductor of capacitance $C$. If charge $+Q$ is given into small amount of $dq$ to the surface of the conductor. Then the work done will be $dw=V\:dq \qquad(1) $ $dw=\frac{q}{C} \:dq \qquad \left (\because V=\frac{q}{C} \right)$ Therefore, as the amount of charge on the conductor will increase from $0$ to $Q$ that causes also increase in work done. So integrate the above equation for total work done $ \int_{0}^{W} dw=\frac{1}{C} \int_{0}^{Q} q dq$ $W=\frac{1}{C} \int_{0}^{Q} q dq$ $W=\frac{1}{C} \left[ \frac{q^{2}}{2} \right]^{Q}_{0}$ $W=\frac{1}{2} \frac{Q^{2}}{C}$ This work is stored in the form of electric potential energy $U$. Then $U=\frac{1}{2} \frac{Q^{2}}{C}$ $U=\frac{1}{2} C V^{2}$ $U=\fr

Derivation → Capacitance of an Isolated Spherical Conductor Let us consider an isolated spherical conductor of radius $a$ is placed in a vacuum or air. Let a charge $+q$ be given to the sphere and this charge is distributed uniformly on the surface of conducting sphere. Then the electric potential at the surface of the conducting sphere is $V=\frac{1}{4\pi \epsilon_{\circ}} \frac{q}{a} \qquad(1)$ So the capacitance of the sphere $C=\frac{q}{V} \qquad(2)$ Now substitute the value of $V$ in equation $(2)$ Then we get $C=\frac{q}{\frac{q}{4\pi \epsilon_{\circ}a}}$ $C=4\pi \epsilon_{\circ}a$ Thus, The capacitance of a spherical conductor is directly proportional to its radius. i.e If the radius of conducting sphere is large then the sphere will hold a large amount of the given charge without running up too high a voltage. Unit: The unit of capacitance is "Farad" i.e. $F$