Explain electric field and also electric field by point charge.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Suppose, charge $\mathrm{Q}$ is placed at origin ' $\mathrm{O}^{\prime}$ in free space. If another charge $q$ is placed at distance $r$ at point $\mathrm{P}(\mathrm{OP}=r)$, then Coulomb force acts on $q$.

$\overrightarrow{\mathrm{F}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\mathrm{Q} q}{r^{2}} \hat{r}$

If $q=1 \mathrm{C}$, then force acting on unit charge is called electric field $\mathrm{E}$.

$\therefore \frac{\overrightarrow{\mathrm{F}}}{q}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\mathrm{Q}}{r^{2}} \hat{r}$ $\therefore \overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\mathrm{Q}}{r^{2}} \hat{r}$ or $\mathrm{E}=\frac{k \mathrm{Q}}{r^{2}}$

Definition of electric field: 'The region around the charge in which the effect of electric charge is prevailing is called the electric field of the charge.'

Electric field $\overrightarrow{\mathrm{E}}$ is also called electric field intensity. Force acting on charge $q$ of position vector $\vec{r}$ is $\overrightarrow{\mathrm{F}}(\vec{r})=q \overrightarrow{\mathrm{E}}(\vec{r})$

Electric field $\vec{E}$ is also called electric field intensity.

Definition of Electric field : 'The force acting on a unit positive charge at a given point in an electric field of a point charge of the system at charge is called the electric field or intensity of electric field $\overrightarrow{\mathrm{E}}$ at that point.

SI unit of electric field intensity is $\mathrm{NC}^{-1}$ or $\mathrm{Vm}^{-1}$ and dimensional formula is $\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$.

897-s113

Similar Questions

Two point charges $Q$ and $-3Q$ are placed at some distance apart. If the electric field at the location of $Q$ is $E$ then at the locality of $ - 3Q$, it is

For the given figure the direction of electric field at $A$ will be

The maximum value of electric field on the axis of a charged ring having charge $Q$ and radius $R$ is

Two charges $ + 5\,\mu C$ and $ + 10\,\mu C$ are placed $20\, cm$ apart. The net electric field at the mid-Point between the two charges is

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \mathrm{~N} / \mathrm{C}$ normally. A charged particle of mass $2 \mathrm{~g}$ being suspended through a silk thread of length $20 \mathrm{~cm}$ and remain stayed at a distance of $10 \mathrm{~cm}$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{\mathrm{x}}} \ \mu \mathrm{C}$ where $\mathrm{x}=$ ____________.  use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]

  • [JEE MAIN 2024]