Physics Vidyapith ✍️

Electric field intensity at different points in the field due to uniformly charged thick hollow non-conducting sphere: Let us consider, A hollow non-conducting sphere of inner radius $r_{1}$ and outer radius $r_{2}$ in which $+q$ charge is evenly distributed evenly in the entire volume of the sphere. If $\rho$ is the volume charge density then electric field intensity at different points on the electric field of the thick hollow non-conducting sphere: Electric field intensity outside the thick hollow non-conducting sphere Electric field intensity on the surface of the thick hollow non-conducting sphere Electric field intensity at an internal point of the non-thick hollow conducting sphere 1. Electric field intensity outside the thick hollow non-conducting sphere: Let us consider, A point $P$ is outside the sphere which is at a $r$ distance from the center point $O$ of the sphere. The direction of electric flux is radially outward in the sphere. So the

A.) Electric field intensity at different points in the field due to the uniformly charged solid conducting sphere: Let us consider, A solid conducting sphere that has a radius $R$ and charge $+q$ is distributed on the surface of the sphere in a uniform manner. Now find the electric field intensity at different points due to the solid-charged conducting sphere. These different points are: Electric field intensity outside the solid conducting sphere Electric field intensity on the surface of the solid conducting sphere Electric field intensity inside the solid conducting sphere 1.) Electric field intensity outside the solid conducting sphere: If $O$ is the center of solid conducting spherical then the electric field intensity outside of the sphere can be determined by the following steps → First, take the point $P$ outside the sphere Draw a spherical surface of radius r which passes through point $P$. This hypothetical surface is known

Electric Potential due to an Electric Dipole: The electric potential due to an electric dipole can be measured at different points: The electric potential on the axis of the electric dipole The electric potential on the equatorial line of the electric dipole The electric potential at any point of the electric dipole 1. The electric potential on the axis of the electric dipole: Let us consider, An electric dipole AB made up of two charges of -q and +q coulomb is placed in a vacuum or air at a very small distance of $2l$. Let a point $P$ is on the axis of an electric dipole and place at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric potential due to dipole's charges. So Electric potential at point $P$ due $+q$ charge of electric dipole→ $ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{r-l}$ The electric potential at point $P$ due $-q$ charge of elect

Electric field intensity due to a uniformly charged infinite plane thin sheet: Let us consider, A plane charged sheet (It is a thin sheet so it will have surface charge distribution whether it is a conducting or nonconducting sheet) whose surface charge density is $\sigma$. From symmetry, Electric field intensity is perpendicular to the plane everywhere and the field intensity must have the same magnitude on both sides of the sheet. Let point $P_{1}$ and $P_{2}$ be the two-point on the opposite side of the sheet. To use Gaussian law, we construct a cylindrical Gaussian surface of cross-section area $\overrightarrow{dA}$, which cuts the sheet, with points $P_{1}$ and $P_{2}$. The electric field $\overrightarrow{E}$ is normal to end faces and is away from the plane. Electric field $\overrightarrow{E}$ is parallel to cross-section area $\overrightarrow{dA}$. Therefore the curved cylindrical surface does not contribute to the flux i.e. $\o

Derivation of electric field intensity due to the uniformly charged wire of infinite length: Let us consider a uniformly-charged (positively charged) wire of infinite length having a constant linear charge density (that is, a charge per unit length) $\lambda$ coulomb/meter. Let P is a point at a distance $r$ from the wire at which electric field $\overrightarrow{E}$ has to find. Let us draw a coaxial Gaussian cylindrical surface of length $l$ through point $P$. By symmetry, the magnitude $E$ of the electric field will be the same at all points on this surface and directed radially outward. Thus, Now take small area elements $dA_{1}$, $dA_{2}$ and $dA_{3}$ on the Gaussian surface as shown in figure below. Therefore, the total electric flux passing through area elements is: $ \phi_{E}= \oint \overrightarrow{E} \cdot \overrightarrow{dA_{1}} + \oint \overrightarrow{E} \cdot \overrightarrow{dA_{2}} + \oint\overrightarrow{E} \cdot \overrightarrow{dA_{3}}$ $

Derivation of electric field intensity due to a point charge by Gauss's Law: Let us consider, a source point charge particle of $+q$ coulomb is placed at point $O$ in space. Let's take a point $P$ on the electric field of the source point charge particle. To find the electric field intensity $\overrightarrow{E}$ at point $P$, first put the test charge particle $+q_{0}$ on the point $P$ and draw a gaussian surface which passes through the point $P$. After that take a very small area $\overrightarrow{dA}$ around the point $P$. If the distance between the source charge particle and small area $\overrightarrow {dA}$ is $r$ then electric flux passing through the small area $\overrightarrow{dA}$ → $ d\phi_{E}= \overrightarrow{E} \cdot \overrightarrow{dA}$ $ d\phi_{E}= E\:dA\: cos\theta$ Electric field due to point charge from the figure, the direction between $\overrightarrow{E}$ and $\overrightarrow{dA}$ is parallel to each other i.e. the angle will be $0^{\ci

Derivation of the electric potential energy of an electric dipole in the uniform electric field: Let us consider an electric dipole $AB$, which is made up of two charges $q_1$ and $q_2$, which are placed at a distance of $2l$ in the electric field $E$. So force acting on each charge due to the electric field will be $qE$. If the dipole gets rotated a small-angle $d\theta$ against the torque acting on it in the uniform electric field $E$ then the small work done is $dW=\tau. d\theta \qquad (1)$ Force of moment on an electric Dipole The torque (i.e moment of force) on an electric dipole in a uniform electric field $ \tau=p.E\:sin\theta$ Now substitute the value of $\tau$ in equation $(1)$. So work done $ dW=p.E\:sin\theta.d\theta$ If the dipole rotate the angle from $\theta_{1}$ to angle $\theta_{2}$ then workdone $\int_{W_{1}}^{W_{2}}dW=p.E\int_{\theta_{1}}^{\theta_{2}}sin\theta \: d\theta$ $ W_{2}-W_{1}=p.E\left[-cos\theta \right]

Principle of Superposition for Electric force: If a system contains a number of interacting charges, then the net force on anyone charge equals the vector sum of all the forces exerted on it by all the other charges. This is the principle of Superposition for electric force . If a system contains n point charges $ q_{1},q_{2},q_{3}........q_{n}$. Then according to the principle of superposition, the force acting on the charge $q_{1}$ due to all the other charges $\overrightarrow{F_{1}}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+...+\overrightarrow{F_{1n}} \qquad (1)$ Where $\overrightarrow{F_{12}}$ is the force on charge $q_{1}$ due to charge $q_{2}$, $\overrightarrow{F_{13}}$ that is due to $q_{3}$ and $\overrightarrow{F_{1n}}$ that due to $q_{n}$. If the distance between the charges $q_{1}$ and $q_{2}$ is $\widehat{r}_{12}$ (magnitude only) and $\widehat{r}_{21}$ is unit vector from charge $q_{2}$ to $q_{1}$, then $\over

Torque on an electric dipole in a uniform electric field: Let us consider, An electric dipole AB, made up of two charges $+q$ and $-q$, is placed at a very small distance $2l$ in a uniform electric field $\overrightarrow{E}$. If $\theta$ is the angle between electric field intensity $\overrightarrow{E}$ and electric dipole moment $\overrightarrow{p}$ then the magnitude of electric dipole moment → $\overrightarrow{p}=q\times\overrightarrow{2l}\qquad(1)$ Force exerted on charge $+q$ by electric field $\overrightarrow{E}$ → $\overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad(2)$ Here in the above equation(2), the direction of $\overrightarrow{F_{+q}}$ is along the direction of $\overrightarrow{E}$ Force exerted on charge $-q$ by electric field $\overrightarrow{E}$ → $\overrightarrow{F_{-q}}= q\overrightarrow{E}\qquad(3)$ Here in the above equation(3) the direction of $\overrightarrow{F_{-q}}$ is in the opposite direction of $\

Electric Dipole: An electric dipole is a system in which two equal magnitude and opposite point charged particles are placed at a very short distance apart. Electric Dipole Moment: The product of magnitude of one point charged particle and the distance between the charges is called the 'electric dipole moment'. It is vector quantity and the direction of electric dipole moment is along the axis of the dipole pointing from negative charge to positive charge. Electric Dipole Let us consider, the two charged particle, which has equal magnitude $+q$ coulomb and $-q$ coulomb is placed at a distance of $2l$ in a dipole so the electric dipole moment is → $\overrightarrow{p}=q\times \overrightarrow{2l}$ Unit: $C-m$ Or $Ampere-metre-sec$ Dimension: $[ALT]$ Electric field intensity due to an Electric Dipole: The electric field intensity due to an electric dipole can be measured at three different points: Electric

Definition of Electric Field Intensity: The force acting on the per unit test charge in electric field is called the Electric field intensity . It is represented by $'E'$. Let us consider that a test-charged particle of $q_{0}$ Coulomb is placed at a point in the electric field and a force $F$ acting on them so the electric field intensity at that point $ \overrightarrow{E}=\frac{\overrightarrow{F}}{q_{0}}$ SI Unit: $\quad Newton/Coulomb$ $ (N/C)$ $\quad Kg-m^{2}/sec^{3} A$ Dimension: $\left [ML^{2}T^{-3}A^{-1} \right ]$ Physical Significance of Electric Field: The force experienced by a charge is different at different points in space. So electric field intensity also varies from point to point. In general, Electric field intensity is not a single vector quantity but it is a set of infinite vector and each point in space have a unique electric field intensity. So electric field is an example of the vector field.

Derivation of Work done by a rotating electric dipole in uniform electric field : Let us consider, An electric dipole AB, made up of two charges $+q$ and $-q$, is placed at a very small distance $2l$ in a uniform electric field $\overrightarrow{E}$. If dipole $AB$ rotates at angle $θ$ from its equilibrium position. If $A'$ and $B'$ are the new position of a dipole in the electric field. Then force on $+q$ charge particle due to electric field→ $ \overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad (1)$ Then force on $-q$ charge particle due to electric field→ $ \overrightarrow{F_{-q}}=q\overrightarrow{E}\qquad(2)$ Work done by rotating an electric dipole   So work done by a force on $+q$ charge particle to bring from position $A$ to position $A'$→ $ \overrightarrow{W_{+q}}=\overrightarrow{F_{+q}}·\overrightarrow{AC}$ $ \overrightarrow{W_{+q}}=q\overrightarrow{E}· \overrightarrow{AC}\qquad(3)$ Similarly, work done by the forc