PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

ELECTRIC POTENTIAL AT ANY POINT DUE TO AN ELECTRIC DIPOLE :  Obtain expression for the electric potential at any point due to an electric dipole. Rewrite this expression if point of observation lies on the (i) axial line of the dipole and (ii) equatorial line of the dipole. Consider any point P at a distance r from the centre (O) of the electric dipole AB. Let OP make an angle $\theta$ with the dipole moment $\vec{p}$. Let $r_1$ and $r_2$ be the distances of point P from -q charge and +q charge of the dipole respectively. Step 1. Potential at point P due to -q charge is given by $V_1 = \frac{1}{4\pi\epsilon_0} \frac{(-q)}{r_1}$ Potential at point P due to +q charge is given by $V_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$ $\therefore$ Using principle of superposition, potential at point P due to the dipole is given by $V = V_1 + V_2$ or $V = -\frac{1}{4\pi\epsilon_0} \frac{q}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$ $V= \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_2} - \frac{1...

Derivation : Capacitance of a Parallel Plate Capacitor with a Dielectric Slab : For a parallel plate capacitor with vacuum (or air) between plates: $C_0 = \frac{\varepsilon_0 A}{d} \quad ...(i)$ where  (A) = area of each plate , (d) = separation between plates , ($\varepsilon_0$) = permittivity of free space . Let a dielectric slab of thickness (t) and dielectric constant (K) be inserted between the plates. - Electric field in free space region (thickness (d - t)): ($E_0$)   - Electric field in dielectric region (thickness (t)): ($E = \frac{E_0}{K}$)   Potential difference between plates: $V = E_0(d-t) + \frac{E_0 t}{K}$ $V= E_0 \left[(d-t) + \frac{t}{K}\right]  ...(ii)$ Since   $E_0 = \frac{q}{A \varepsilon_0}$ we get   $V = \frac{q}{A \varepsilon_0} \left[(d - t) + \frac{t}{K}\right] \quad ...(iii)$ By definition, $C = \frac{q}{V}$ $C = \frac{q}{ \frac{q}{A \varepsilon_0} \left[(d - t) + \frac{t}{K}\right]}$ $C= \frac{\varepsilon_0 A}{...

Derivation: Capacitance of a Parallel Plate Capacitor Consider a parallel plate capacitor consisting of two parallel plates separated by a distance ( d ).   - Each plate has an area ( A ).   - Plate R carries charge ( +q ), and Plate S carries charge ( -q ).   - The surface charge densities are ( $+\sigma$ ) and ( $-\sigma$) respectively.   The electric field between the plates is: $E = \frac{\sigma}{\varepsilon_0} \quad ...(i)$ where ( $\varepsilon_0$ ) is the permittivity of free space (assuming the medium is air or vacuum).   The direction of the uniform electric field is from the positive plate to the negative plate. Derivation :  - Potential difference between the plates: $V = E \cdot d$ - Since ( $\sigma = \frac{q}{A}$ ), substituting into equation (i): $E = \frac{q}{A \varepsilon_0}$ So, $V = \frac{q}{A \varepsilon_0} \cdot d$ - Capacitance is defined as: $C = \frac{q}{V}$ Substituting ( V): $C = \frac{q}{\frac{q d}{A \vareps...

Notes : Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance - Physics Kund Chapter 2: Electrostatic Potential and Capacitance Electrostatic Potential Energy Electric Potential Potential Difference Electric Potential Due to a Point Charge Electric Potential at Any Point Due to an Electric Dipole Electric Potential Due to a Group or System of Point Charges (Principle of Superposition of Potentials) Equipotential Surfaces Electric Potential Energy of a System of Two and Three Point Charges Expression for Potential Energy of Single and Two Charges in an External Electric Field Electric Potential Energy of an Electric Dipole in an Electric Field Electrostatic Shielding and Its Significance Behaviour of Conductors in an Electrostatic Field Capacitors and Capacitance Capacitance of a Parallel Plate Capacitor Effect of Dielectric on Capacitance Combination of Capacitors in Series Combination of Capacitors in ...

ENERGY STORED IN A CHARGED CAPACITOR (ELECTROSTATIC ENERGY IN A CAPACITOR) Derive an expression for the energy stored in a charged capacitor. In what form the energy is stored in a capacitor? The process of charging a capacitor is equivalent to that of transferring charge from one plate to the other plate of the capacitor. At any stage of the charging, there is a potential difference between the plates of the capacitor. Therefore, some work must be done to transfer charge from one plate to another plate of the capacitor. This work done is stored as electrostatic potential energy in the capacitor. Let at any instant, a charge $q$ be on the capacitor. Then potential difference between the plates of the capacitor is given by $V = q/C$. If extra charge $dq$ is transferred to the capacitor, then work done to do so is given by $dW = V dq = \frac{q}{C} dq \quad \dots (i)$ If the final charge on the capacitor is $Q$, then the total work done is given by $W = \int dW = \int_0^Q \frac{q}{C} dq =...

Question: What do you understand by capacitors in parallel ? Derive a relation for equivalent capacitance of n capacitors connected in parallel. Capacitors in Parallel: Two or more capacitors are said to be connected in parallel if left plates of all capacitors are connected to one terminal and the right plates of all capacitors are connected to other terminal of the battery such that the potential difference across each capacitor is same. Expression for Equivalent Capacitance : Let $q_1$ and $q_2$ be the maximum charges on $C_1$ and $C_2$ respectively. The total charge $q$ on the system of two capacitors connected in parallel is given by $q = q_1 + q_2$ But $q_1 = C_1 V \quad \text{and} \quad q_2 = C_2 V$ $q = C_1 V + C_2 V = (C_1 + C_2) V \quad \dots \text{(i)}$ If C = Capacitance of parallel combination of the capacitors (called equivalent capacitance of the combination) $q = CV$ then $CV = (C_1 + C_2) V$ Hence eqn. (i) becomes $C = C_1 + C_2$ Which is the equivalent capacitance of ...

Question: What do you understand by capacitors in series ? Derive a relation for equivalent capacitance of n capacitors connected in series. Answer : Two or more capacitors are said to be connected in series if the right plate of one capacitor is connected to the left plate of other capacitor such that the magnitude of the charge (q) on each capacitor is same, when connected to a cell or battery. Expression for Equivalent Capacitance: Two capacitors of capacitance $C_1$ and $C_2$ connected in series. Let $V$ be the applied potential difference (through battery) across the two terminals A and B of the series combination of capacitors. Total potential difference $V$ across the arrangement will be divided as $V_1$ and $V_2$ across each capacitor i.e. $C_1$ and $C_2$ respectively. When capacitors are connected in series, potential difference across the combination is the sum of the potential differences across each capacitor. i.e., $V = V_1 + V_2  \quad \dots \text{(i)}$ But  $V_1...

CAPACITOR OR CONDENSER Define a capacitor. Give symbol, types, purpose and uses of capacitor. Capacitor: A capacitor consists of two conductors of any shape separated by a non-conducting medium (insulator or dielectric) such that it can store electric charge. When the capacitor is charged by connecting the two uncharged conductors to the terminals of a battery, the two conductors carry charges of equal magnitude but of opposite sign. Now, when the battery is disconnected, the charges on the conductors remain the same. Hence, the capacitor stores the charge. The potential difference ($V = V_1 - V_2$) between the two conductors is the potential difference across the capacitor. This potential difference (V) is proportional to the charge (q) on the capacitor. That is, $V \propto q$ $q = CV$ ... (1) where C is known as the capacitance of the capacitor. Therefore, $C = \frac{q}{V}$ ... (2) Thus, capacitance (C) of a capacitor is defined as the ratio of the magnitude of the charge q on the c...

Electrostatics of Conductors (Conductors In Electrostatic Field)} Discuss the behaviour of an conductor in an electrostatic field. Derive expression for electric field at the surface of a charged conductor. Behaviour of Conductors in the Electrostatic Field 1. Net electrostatic field inside a conductor is zero. A conductor contains large number of free electrons. When it is placed in the electrostatic field ($\vec{E}_0$), each electron experiences a force ($\vec{F} = -e\vec{E}_0$) in a direction opposite to the direction of applied field $\vec{E}_0$. This force causes free electrons in the conductor to move in a direction opposite to the direction of the applied electric field $\vec{E}_0$. Therefore, side A of the conductor becomes positively charged due to the transfer of free electrons to the side B and the side B of the conductor becomes negatively charged. In other words, side A of the conductor has less electrons and side B of the conductor has more electrons. This re-distribution...

What do you understand by electrostatic shielding? Explain its significance. Electrostatic Shielding : The method of protecting a region from the effect of external electric field is called electrostatic shielding. When a conductor having a cavity of any shape is placed in an external electric field, the charges of conductor re-distribute and resides on the outer surface of the conductor as shown in figure. The electric field inside the conductor including cavity is zero. Thus, the cavity in the conductor will not experience any force due to the charges on the outer surface of the conductor. This is called electrostatic shielding. The concept of electrostatic shielding is used to protect sensitive instruments from the effect of strong electrostatic fields by enclosing these instruments in a metallic cage. It is for this reason that it is safer to sit in a car or a bus during lightning and thunder rather than to stand under a tree or on the open ground.

Electric Potential Energy of an Electric Dipole in an Electric Field Derive an expression for the electric potential energy of an electric dipole placed in a uniform electric field. Let an electric dipole of dipole moment $\vec{p}$ be placed in an electric field $\vec{E}$ making an angle $\theta$ with the direction of electric field intensity $\vec{E}$. The torque acting on the dipole is given by, $ \tau = pE\sin\theta \qquad \cdots (i)$ Work done to rotate the dipole through an angle $d\theta$ is given by, $dW = \tau d\theta = pE\sin\theta d\theta$ Work done in rotating the dipole from an angle $\theta_1$ to $\theta_2$ is given by, $W = \int dW = \int_{\theta_1}^{\theta_2} pE\sin\theta d\theta$ $W= pE\int_{\theta_1}^{\theta_2} \sin\theta d\theta$ If $\theta_1 = 90^\circ$ and $\theta_2 = \theta$, then     $W= pE [-\cos \theta]_{\theta_1}^{\theta_2} = -pE [\cos \theta_2 - \cos \theta_1]$     $= -pE [\cos \theta - \cos 90^\circ]$     $= -pE \cos \theta \qqu...

POTENTIAL ENERGY OF CHARGES IN AN EXTERNAL ELECTRIC FIELD  Derive an expression for the potential energy in the cases of (i) single point charge and (ii) system of two charges in an external electric field. Potential Energy of a Single Charge in an External Electric Field :  Let an external field $\vec{E}$ have different values of electric potential at different points. Consider a point P, distant $\vec{r}$ from the origin in this field having electric potential as $V(\vec{r})$. Then, work done in bringing charge $q$ from infinity to point P is given by,  $W = qV(\vec{r})$. This work done is stored as potential energy of the charge $q$. Thus, potential energy of charge $q$ at position vector $\vec{r}$ in external field is given by, $U = qV(\vec{r})$ Potential energy of a system of two charges in an external electric field. Let $q_1$ and $q_2$ be two charges placed at points P and Q having position vectors $\vec{r}_1$ and $\vec{r}_2$ respectively in an external field $\vec...

Electric Potential Energy of a System of Two and Three point charges Define electric potential energy of a system of charges. Derive expressions for potential energy of a system of (i) two point charges, (ii) three point charges and (iii) n point charges. Electric potential energy of a system of point charges is defined as the total work done in bringing these point charges to their respective locations from infinity to form a system of charges. Potential Energy of a System of Two Point Charges Consider two point charges $q_1$ and $q_2$ initially lying at infinity. Work done to bring a charge $q$ from infinity to a point, where electric potential due to any source charge is $V(r)$ is given by $ W = q V(r)$ When charge $q_1$ is brought from infinity to a position A, no work is done as $V(r) = 0$ at A in the absence of any source charge  When charge $q_2$ is brought from infinity to the position B, then the work done is given by $W = q_2 \times V_1 \qquad ...(1)$ where, $V_1 = \frac{...

Relation between Electric Field Intensity and Potential (or $E = -\frac{dV}{dr}$) Show that electric field intensity is given by the negative gradient of electric potential. What does the negative sign indicate in this expression? Consider two points A and B in the electric field $\vec{E}$ due to a point charge $+Q$ placed at O. Assume that points A and B are very close to each other so that electric field intensity between A and B is uniform. Let  $q_0$  be the positive test charge placed at A. The force acting on charge  $q_0$  in the electric field  $\vec{E}$  is given by  $\vec{F} = q_0 \vec{E}$.  The direction of force $\vec{F}$ is along the direction of $\vec{E}$ i.e., away from the charge  $+Q$ . Work done to move the test charge from point A to point B through distance $(dr)$ is given by :  $dW= \vec{F} \cdot d\vec{r}$ $(\because \vec{F} = q_0 \vec{E})$ $dW= q_0 \vec{E} \cdot d\vec{r}$ $(\because \cos 180^\circ = -1)$ $dW= q_0 E ...

EQUIPOTENTIAL (SAME POTENTIAL) SURFACE  Define equipotential surface. List the factors on which formation of equipotential surface depends. Draw and explain equipotential surfaces for (i) uniform electric field, (ii) isolated point charge, (iii) a pair of similar point charges and (iv) electric dipole. Equipotential Surface : A surface whose every point has same electric potential due to charge configuration is called equipotential surface. An equipotential surface is defined as the locus of all the points in a medium at which electric potential due to a charge configuration is same. The formation of an equipotential surface will depend upon the type of medium i.e., isotropic or non-isotropic and the amount of charge distribution. 1. Equipotential surfaces for a uniform electric field Uniform electric field is represented by equidistant parallel straight lines. Draw planes I, II and III perpendicular to the direction of electric field (Figure 17). The potential $V_1$ at every point...

2.4. ELECTRIC POTENTIAL DUE TO A GROUP OR SYSTEM OF POINT CHARGES (PRINCIPLE OF SUPERPOSITION OF POTENTIALS) State principle of superposition of electric potentials and use it to find the expression for the electric potential at a point due to a group of point charges. Principle of superposition of potentials states that the net potential at any point due to $n$ discrete charges is given by the algebraic sum of their individual potentials at that point. Consider $n$ discrete positive point charges $q_1, q_2, q_3, \dots, q_n$ at distances $r_1, r_2, r_3, \dots, r_n$ respectively from a point P. Potential at P due to charge $q_1$, $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1}$ Potential at P due to charge $q_2$, $V_2 = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2}$ Potential at P due to charge $q_n$, $V_n = \frac{1}{4\pi\epsilon_0} \frac{q_n}{r_n}$ Applying principle of superposition of potentials for a group of charges, we get net electric potential V at P due to n charges $ V = V_1 + V_2 ...

2.3. ELECTRIC POTENTIAL DUE TO A POINT CHARGE Derive an expression for electric potential due to a point charge. What is the nature of symmetry of electric potential due to single charge and why? Consider a point charge q. Let a unit positive charge (+1) be placed at a point P in the electric field of the charge q. Let the distance between +q and unit positive charge (+1) be x. The electrostatic or Coulomb's force of repulsion between charge $q$ and +1 charge is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q \times 1}{x^2}$ or $F = \frac{1}{4\pi\epsilon_0} \frac{q}{x^2}$...(1) This force is directed radially outwards along the line joining +q charge and +1 charges (i.e., in the direction of $\vec{E}$)  Now, let unit positive charge be displaced towards +q charge, without acceleration, such that the displacement of unit positive charge is $dx$. Work done in displacing unit positive charge through $dx$ is given by $dW = \vec{F} \cdot d\vec{x}$ $dW= F dx \cos 180^\circ = -F dx$  ...

Potential Difference State and find a relation for potential difference. Potential difference: Potential difference between any two points in an electric field is defined as the work done in moving a unit positive charge from one point to the other point against the electric force of the electric field irrespective of the path followed. Potential difference between any two points in an electric field is defined as the electric potential energy difference per unit charge. Let $\Delta U = U_B - U_A$ be the electrostatic potential energy difference between two points A and B in the electric field. The electric potential difference $\Delta V$ between points A and B is given by $\Delta V = V_B - V_A = \frac{\Delta U}{q_0}$ $\Delta V = \frac{U_B - U_A}{q_0} = \frac{W_{AB}}{q_0}$ Now, $U_B - U_A = -q_0 \int_A^B \vec{E} \cdot d\vec{r}$ $\Delta U = -q_0 \int_A^B \vec{E} \cdot d\vec{r}$ \Delta V = V_B - V_A = \frac{\Delta U}{q_0}$ $= \frac{-q_0 \int_A^B \vec{E} \cdot d\vec{r}}{q_0} = -\int_A^B \...

ELECTRIC (OR ELECTROSTATIC) POTENTIAL  What do you understand by electric potential Explain. Give dimensional formula and SI unit of electric potential. What is the unit of electric potential in CGS system? Establish the relation between SI and CGS units of electric potential. Electric potential: Electric potential at a point in the electric field is defined as the work done in moving a unit positive test charge from infinity to that point in the electric field of the source charge. Expression for Electric potential: Electric potential at a point in an electric field is also defined as the electric potential energy per unit charge. Let U be the electrostatic potential energy at a point (say B) in the electric field. Then, the electric potential at that point in the electric field is given by $V = \frac{U}{q_0} = \frac{W}{q_0}...(1)$ But electric potential energy of test charge $q_0$ at point (say B) in the electric field is given by $U = -q_0 \int_{\infty}^B \vec{E} \cdot d\vec{r}$...

2.1. ELECTRIC OR ELECTROSTATIC POTENTIAL ENERGY 1. What is electric or electrostatic potential energy? Derive an expression for electric potential energy. Electrostatic potential energy: Electric potential energy of a charge at a point in the electric field due to another charge is defined as the work done by an external force in bringing a test charge (without acceleration) from infinity to that point in the electric field. Let a positive test charge $q_0$ be placed at A in an electric field $\vec{E}$ due to another positive charge Q (called source charge) as shown in figure 1. The force acting on the test charge $q_0$ in the electric field $\vec{E}$ is given by $\vec{F}_e = q_0 \vec{E}$ This force tends to move the test charge in the direction of the electric field $\vec{E}$ (i.e., away from +Q charge). Suppose an external force $\vec{F}_o$ acts just to overcome the electric force $\vec{F}_e$ on the test charge to move it without any acceleration towards source charge Q. If the test ...