PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Coulomb's Law in Vector Form :  Question: Write Coulomb's law in vector form. Show that Coulomb's law agrees with Newton's third law of motion ? Consider two point charges $q_1$ and $q_2$ of same polarity separated by a distance $r$ (Figure 16). Coulomb's force acting on $q_1$ due to $q_2$ is given by, $\vec{F}_{12} = k \frac{q_1 q_2}{r^2} \hat{r}_{21} \quad \dots (i)$ where $\hat{r}_{21}$ is the unit vector directed from $q_2$ to $q_1$. Thus, Coulomb's force acting on point charge $q_1$ due to point charge $q_2$ is directed along the line joining $q_1$ and $q_2$ from charge $q_2$ to charge $q_1$. Similarly, force acting on $q_2$ due to $q_1$ is given by $\vec{F}_{21} = k \frac{q_1 q_2}{r^2} \hat{r}_{12} \quad \dots (ii)$ where $\hat{r}_{12}$ is the unit vector directed from $q_1$ to $q_2$. Thus, Coulomb's force acting on point charge $q_2$ due to point charge $q_1$ is directed along the line joining $q_1$ and $q_2$ from charge $q_1$ to charge $q_2$. Eqns. (...

According to the superposition principle}, the net force acting on a given point charge due to a number of point charges around it is the vector sum of the individual forces acting on that point charge due to all other point charges. Consider number of point charges $q_1, q_2, q_3, \dots, q_n$ having position vectors $\vec{r}_1, \vec{r}_2, \vec{r}_3, \dots, \vec{r}_n$ respectively. Let $q_0$ be the test charge having position vector $\vec{r}_0$. Let $\vec{F}_{01}, \vec{F}_{02}, \vec{F}_{03}, \dots, \vec{F}_{0n}$ be the forces acting on the given test charge $q_0$ due to $q_1, q_2, q_3, \dots, q_n$ respectively, then the net force acting on $q_0$ is given by $\vec{F} = \vec{F}_{01} + \vec{F}_{02} + \vec{F}_{03} + \dots + \vec{F}_{0n}$ $\vec{F} = \sum_{i=1}^{n} \vec{F}_{0i} \quad \dots (i)$ The force acting on $q_0$ due to $q_1$ is given by $\vec{F}_{01} = \frac{1}{4\pi\epsilon_0} \frac{q_0 q_1 (\vec{r}_0 - \vec{r}_1)}{|\vec{r}_0 - \vec{r}_1|^3} \quad \dots (ii)$ The force acting on $q_0...

सतत आवेश वितरण (Continuous Charge Distribution) परिभाषा जब आवेश किसी वस्तु पर असंख्य छोटे-छोटे भागों में फैला हो (न कि एक या कुछ बिंदुओं पर), तो उसे सतत आवेश वितरण कहते हैं। > जैसे — किसी तार, चादर या ठोस गोले पर फैला आवेश सतत आवेश वितरण के प्रकार 1. रेखीय आवेश वितरण (Linear Charge Distribution) जब आवेश किसी रेखा या तार पर फैला हो। रेखीय आवेश घनत्व (λ): $\lambda = \frac{dq}{dl}$ $\Rightarrow dq = \lambda \cdot dl$ - मात्रक (Unit):C/m (कूलॉम प्रति मीटर) - उदाहरण: आवेशित तार, छड़ > कुल आवेश: $Q = \int \lambda \, dl$ 2. पृष्ठीय आवेश वितरण (Surface Charge Distribution) जब आवेश किसी पृष्ठ (surface) पर फैला हो। पृष्ठीय आवेश घनत्व (σ): $\sigma = \frac{dq}{dA}$ $\Rightarrow dq = \sigma \cdot dA$ - मात्रक (Unit): C/m² (कूलॉम प्रति वर्ग मीटर) - उदाहरण: आवेशित चादर, गोले की सतह > कुल आवेश: $Q = \int \sigma \, dA$ 3. आयतनिक आवेश वितरण (Volume Charge Distribution) जब आवेश किसी आयतन (volume) में फैला हो। आयतनिक आवेश घनत्व (ρ): $\rho = \frac{dq}{dV}$ $\Rightarrow dq = \rho \cdot dV$ - मात्रक...

Electric Field Intensity due to a Point Charge Question: Obtain an expression for electric field intensity due to a point charge (i) when source charge is at the origin, (ii) when source charge is away from origin. Case 1. When source charge is at the origin. Consider a static point charge Q at the origin O (0, 0, 0). Let $q_0$ be the test charge placed in free space at a point P, distant r from O in the electric field of the charge Q. According to Coulomb's law, force experienced by $q_0$ due to charge Q is given by $\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{r^2} \hat{r}$...(i) where $\hat{r}$ is a unit vector directed along OP. Now, electric field intensity at point P due to point charge Q is given by $\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{q_0} \times \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{r^2} \hat{r}$ $\vec{E}= \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \hat{r}...(ii)$ Magnitude of the electric field intensity is given by $E = |\vec{E}| = \frac{1}{4\pi \epsilon_0} \frac{Q}...

Define electric field intensity. Is it a vector or scalar quantity? Give S.I. unit and dimensional formula of electric field intensity. Electric Field Intensity: The electric field intensity due to a static point charge at any point in its electric field is defined as the force experienced by a unit positive charge (i.e., test charge) placed at that point. Let a test charge $q_0$ be placed at a point P in the electric field of the static point charge Q. If $\vec{F}$ is the force experienced by the test charge $q_0$ in the electric field of the point charge, then electric field intensity of the point charge at point P is given by $\vec{E} = \frac{\vec{F}}{q_0} ...(1)$ where $q_0 \rightarrow 0$ (i.e., $q_0$ is infinitesimally small) so that presence of this charge may not disturb the location of the point source charge producing the electric field. . Hence, expression for electric field intensity can be written as $\vec{E} = \underset{q_0 \to 0}{\text{Limit}} \frac{\vec{F}}{q_0} ...(2)$ ...

Definition of Coulomb's law:  According to Coulomb's law, the magnitude of force of attraction or repulsion between any two point charges at rest is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them. Let $r$ be the distance between two static point charges $q_1$ and $q_2$. Then according to Coulomb's law, the force between charges is given by $F \propto q_1 q_2$ $F \propto \frac{1}{r^2}$  or $F \propto \frac{q_1 q_2}{r^2}$ $F = \frac{k q_1 q_2}{r^2} \quad \dots (1)$ where $k$ is the constant of proportionality, known as electrostatic force constant or Coulomb's constant. The value of electrostatic force constant ($k$) depends upon (i) the system of unit used and (ii) the nature of medium in which charges are placed. Coulomb's Law in SI :  In SI, $k = \frac{1}{4\pi \epsilon} = \frac{1}{4\pi \epsilon_0 \epsilon_r}$,  where, $\epsilon_0 = 8.854 \times 10^{-12} \text{C}^2 \text{N}^{-1} ...

Electric Field Intensity due to a System of Charges (Principle of Superposition) Question: Find an expression for electric field intensity at a point due to a system of point charges. Principle of Superposition of Electric Fields: According to the principle of superposition, net electric field strength at a point due to a group of point charges is equal to the vector sum of all the electric field strengths produced due to individual point charges at that point. Suppose we have $n$ point charges $q_1, q_2, \ldots, q_n$ such that their position vectors are $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n$ respectively. Let $q_0$ be the positive test charge at point $P$ where the total electric field $\vec{E}$ due to $n$ charges is to be determined. Let the position vector of the point $P$ be $\vec{r}$ . Electric fields due to point charges $q_1, q_2, \ldots, q_n$ respectively at point $P$ are given by, $\vec{E}_1= \frac{1}{4\pi\epsilon_0} \frac{q_1(\vec{r} - \vec{r}_1)}{|\vec{r} - \vec{r}_1|^3} ...

Electric field line : Electric field line is a straight or curved imaginary line in a region such that the tangent at any point in the field line gives the direction of the electric field at that point in the region.  The theoretical concept of electric field lines was introduced by Faraday. (i) Electric field lines for a single static positive point charge have direction radially outwards. (ii) Electric field line due to a static  negative point charge have direction radially inwards. (iii) Electric field line for two equal postive point charges : Pont N represents the neutral point , where net electric field intensity due to both charges is zero and no electric field line passes through this point. Neutral point lies at the exact middle of the distance two similar and equal charges. Type of electric field :  (i) Uniform electric field : An electric field which has same magnitude and direction at every point in a region is called uniform electric field. It is represented...

Definition of Electric dipole :  A pair of two equal and opposite charges seperated by a small distance is called an electric dipole. Total charge on electric dipole = +q + (-q) = 0. However the charge on electric dipole is equal to the magnitude of either charge of the electric dipole  Examples of electric dipoles : Molecules of water $(H_{2}O)$ , ammonia $(NH_{3})$ , hydrochloric acid (HCL), carbon dioxide $(CO_{2})$, Sodium chloride (NaCl) , alcohal etc  Electric Dipole Moment :  Electric dipole moment of an electric dipole is defined as the product of the magnitude of either charge of the electric dipole and the dipole length.  $$\vec{p} = q(2 \vec{l})$$ The magnitude of the dipole moment is  $$p = q \times 2l$$ The direction of dipole moment $\vec{p}$ is from negative to positive charge. Dipole moment is a vector quantity. S.I unit of dipole moment : $(\vec{p})$ is Coulomb metre (C-m) Dimensional formula of dipole moment : $$p = q \times 2l$$ $$p = I.t...

Question : What is meant by 'continuous charge distribution'? Define: linear charge density, surface charge density and volume charge density. Write their units in SI. Find Force and Field due to continuous charge distribution  Continuous charge distribution A system of closely spaced electric charges forms a continuous charge distribution. At macroscopic scale, the number of charges is very large. Practically at this level, it is not feasible to count the number of charges, so quantization of charges can be easily ignored. In other words, it can be said that at macroscopic level, the charges are supposed to be continuously distributed. (i) Linear charge density (λ) : Linear charge density is defined as charge per unit length. If a charge q is uniformly distributed along a line or a wire of length l, then the linear charge density of the line or the wire is given by: $\lambda = \frac{q}{l}$ The SI unit of $\lambda$ is coulomb/metre (C m$^{-1}$). The charge on a small element of...

GAUSS'S THEOREM OR GAUSS'S LAW :  Question: State and explain Gauss' theorem for electrostatic. Solution :  Gauss's Theorem: According to Gauss's theorem, the total electric flux ($\phi$) through any closed surface (S) in free space is equal to $1/\epsilon_0$ times the total electric charge ($q$) enclosed by the surface. $\phi = \oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0}$  $\phi = \frac{q}{\epsilon_0}$ Proof: Consider an isolated point charge $+q$ placed at point $O$. Let surface $S$ be a sphere of radius $r$ around the charge $+q$. This sphere is called Gaussian surface. What is Gaussian Surface ?  Gaussian surface around a charge (point or continuous distribution) is an imaginary closed surface, such that the intensity of electric field at all points on its surface is same.} Proof of Gauss's Theorem or Gauss's law :   Electric field intensity due to charge $+q$ at every point on the Gaussian surface is given by : $\vec{E} = \frac{1}{4\pi\epsilon...

Question: What is electric flux ? Give an expression for electric flux. Give S.I unit and dimensional formula for electric flux. Solution:  Electric Flux: Electric flux linked with any surface is defined as the total number of electric field lines passing through that surface. Consider a surface of area S. Let electric field lines representing electric field $\vec{E}$ pass normally through the plane of area S. Then the electric flux through the surface of area S is given by $\phi = ES \qquad \dots(i)$ Thus, electric flux may be defined as the product of the magnitude of the electric field (E) and the surface area (S) perpendicular to the electric field. Let $\theta$ be the angle made by the electric field $\vec{E}$ with the area vector $d\vec{S}$ of the surface element. (E cos $\theta$) is the component of the electric field perpendicular to the surface area. Then, the electric flux through the surface element is given by :  $d\phi = (E \cos \theta) dS$ or $d\phi = E dS \cos \...

Question: Use Gauss's theorem to find electric field intensity due to an infinitely long straight Uniformly charged thin wire or conductor. Solution:  Consider an infinite and very thin straight uniformly charged wire having linear charge density (i.e., charge per unit length) $\lambda$. To calculate the electric field intensity $E$ at a point $P$, distant $r$ from the line, draw an imaginary cylinder (Gaussian surface) of radius $r$ and length $l$ around the charged line. The charge enclosed by the Gaussian surface, \[ q = \lambda l \] According to Gauss' theorem, \[ \oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0} = \frac{\lambda l}{\epsilon_0} \qquad \text{...(1)} \] The cylindrical Gaussian surface is divided into three parts I, II and III i.e. top circular face, bottom circular face and curved surface respectively as shown in figure 63. Therefore, eqn. (1) can be written as, $\oint \vec{E} \cdot d\vec{S} =$ $=\int_I \vec{E} \cdot d\vec{S} + \int_{II} \vec{E} \cdot d\v...

Question : Use Gauss'theorem to find electric field intensity due to a Uniformly charged infinite plane sheet of charge . Solution: Consider a thin infinite plane sheet having uniform surface charge density (charge per unit area) $\sigma$ To calculate the electric field intensity $\vec{E}$ at a point P , distant r from the sheet , draw a Gaussian surface in the form of a closed cylinder of length r on each side of sheet with end caps of area S. Electric field $\vec{E}$ is perpendicular to the sheet and hence it is perpendicular to the planes of ends caps I and II also. Charge enclosed by Gaussian surface  $q = \sigma S$ According to Gauss'theorem  $\oint_{S} \vec{E} \cdot d\vec{S}= \frac{q}{\epsilon_{0}}$ $\oint_{S} \vec{E} \cdot d\vec{S}= \frac{\sigma S}{\epsilon_{0}}$ .....(i) Gaussian surface is divided into three parts I, II, III i.e. end caps and the curved surface of the cylinder  From eqn. (i) becomes $\oint_{I} \vec{E} \cdot d\vec{S} + \oint_{II} \vec{E} \cdot d\v...

Question: Using Gauss's theorem , derive expression for electric field intensity due to a Uniformly Charged hollow sphere (a) at a point outside the shell (b) at a point on the surface of the shell (c) at a point inside the shell Solution: (a) Electric field intensity at a point outside the shell :  Consider a positive charge q distributed uniformly on the surface of a spherical shell of Radius R. Let P is a point outside the shell at a distance r from the centre of the shell ( r>R) , where electric field is to be calculated. According to Gauss's theorem      $\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_{0}}$    $ \oint_{S} E \, dS \cos \theta = \frac{q}{\epsilon_{0}}$ ....(i) Since $\vec{E}$ and $d\vec{S}$ are along the same direction, so $\theta = 0^\circ$.  $\therefore \oint_{S} E \, dS \cos 0^\circ = \frac{q}{\epsilon_{0}}$ or  $\oint_{S} E \, dS = \frac{q}{\epsilon_{0}}$ Since E at all points on the Gaussian surface is same and di...

Electric Dipole in an External Uniform Electric Field ( Torque on a Dipole in uniform Electric Field) Consider an electric dipole in a uniform electric field $\vec{E}$ such that the dipole moment ($\vec{p}$) of the electric dipole makes an angle $\theta$ with the electric field $\vec{E}$  Charge +q and -q constituting the dipole , experience equal and opposite forces given by $\vec{F} = q \vec{E}$ and $\vec{F}= -q \vec{E}$ respectively due to electric field $\vec{E}$ Therefore , the net force on the electric dipole is given by $\vec{F_{net}} = q \vec{E} - q\vec{E} = 0$ This , Net force on an electric dipole in a uniform electric field is zero. Therefore , electric dipole placed in uniform electric field does not undergo any translatory motion.In other word , electric dipole is in translational equilibrium in an uniform electric field. But two equal and opposite forces acting on the dipole constitute a couple. This couple tends to rotate the dipole in the clockwise direction and hen...

Question : Derive an expression for Electric Field intensity at a point on the equatorial line of an electric dipole ( Broad side on Position) Solution : Consider an electric dipole AB of length $2l$. Let P be the point on the equatorial line at a distance r from the centre O of the dipole. Step 1. Electric field intensity at P due to +q charge is given by $\vec{E}_{1} = \frac{1}{4\pi\epsilon_{0}} \frac{|+q|}{BP^2}$ along BA $\vec{E}_{1}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(\sqrt{r^2 + l^2})^2}$ $\vec{E}_{1}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$ along BA ...(i) Electric field intensity at P due to $-q$ charge is given by,  $\vec{E}_{2} = \frac{1}{4\pi\epsilon_{0}} \frac{|-q|}{AP^2}$along BA $\vec{E}_{2}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(\sqrt{r^2 + l^2})^2}$ $\vec{E}_{2}= \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$along BA ....(ii) From (1) and (2), $|\vec{E}_{1}| = |\vec{E}_{2}| = \frac{1}{4\pi\epsilon_{0}} \frac{q}{(r^2 + l^2)}$ ...(iii) $\vec{E}_{1}$ and ...

Questions: Derive an Expression for Electric Field Intensity at a Point on the Axial line of An Electric Dipole (End on Position) when the point is on the dipole axis Solution:  Consider an electric dipole consisting of charges +q and -q. The length of the dipole is 2l. Consider a point P at a distance $r$ from the centre O of the dipole on the axial line of the dipole. {STEP 1.} Electric field intensity at P due to $+q$ charge is given by :  $\vec{E}_{+q} = \frac{1}{4\pi\epsilon_0} \frac{q}{BP^2} \quad \text{along (+) x-axis}$ $ = \frac{1}{4\pi\epsilon_0} \frac{q\hat{i}}{(r-l)^2}$ Electric field intensity at P due to $-q$ charge is given by $\vec{E}_{-q} = \frac{1}{4\pi\epsilon_0} \frac{q}{AP^2} \quad \text{along (-) x-axis} $ $= \frac{1}{4\pi\epsilon_0} \frac{q(-\hat{i})}{(r+l)^2}$ $= \frac{-1}{4\pi\epsilon_0} \frac{q\hat{i}}{(r+l)^2}$ {STEP 2.} Net electric field intensity at point P due to electric dipole is given by $\vec{E} = (\vec{E}_{+q}) + (\vec{E}_{-q})$ $= \lef...

NCERT Solutions For Class 12 Physics Chapter 1 Electric Charges And Fields is prepared and uploaded for reference by academic team of expert members Question 1.1 :  What is the force between two small charged spheres having charges of 2 × 10 -7 C and 3 × 10 -7 C placed 30 cm apart in air? Solution :  \[F = 9×10^{9} \frac{q_{1}q_{2}}{r^{2}}\] \[F = \frac{9×10^{9}×2×10^{-7}×(3×10^{-7)}}{0.30×0.30}\] \[F = 6×10^{-3} N\] Question 1.2 :   The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Solution : (a) Force on charge 1 due to charge 2 is given by the relation \[F_{12} = 9×10^{9}\frac{q_{1}q_{2}}{r^{2}}\] \[r^{2} = \frac{(9×10^{9})q_{1}q_{2}}{F_{12}}\] \[r^{2} =\frac{(9×10^{9})(.8×10^{-6})(.4×10^{-6})}{.2}\] \[r^{2} =\frac{2.88×10^{-3}}{2×10^{-1}}\] \[r^{2}= 1.44 × 10^{-2}\] \[r = \sqrt{1.44 × 10...