Explain the concept of an electric field and derive the expression for the electric field due to a point charge.
(N/A) Definition of Electric Field: The region surrounding an electric charge in which its influence can be experienced by another charge is called the electric field of that charge.
Electric Field due to a Point Charge:
Suppose a point charge $Q$ is placed at the origin $O$ in free space. If a test charge $q$ is placed at a distance $r$ at point $P$ (where $OP = r$),then according to Coulomb's law,the force acting on $q$ is:
$\overrightarrow{F} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q q}{r^{2}} \hat{r}$
The electric field $\overrightarrow{E}$ at a point is defined as the force experienced by a unit positive test charge placed at that point:
$\overrightarrow{E} = \frac{\overrightarrow{F}}{q}$
Substituting the expression for force:
$\overrightarrow{E} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{r^{2}} \hat{r}$ or $E = \frac{k Q}{r^{2}}$
Key Properties:
$1$. Electric field $\overrightarrow{E}$ is also known as electric field intensity.
$2$. The force acting on a charge $q$ at position vector $\vec{r}$ is given by $\overrightarrow{F}(\vec{r}) = q \overrightarrow{E}(\vec{r})$.
$3$. The $SI$ unit of electric field intensity is $N C^{-1}$ or $V m^{-1}$.
$4$. The dimensional formula is $[M^{1} L^{1} T^{-3} A^{-1}]$.
Electric Field due to a Point Charge:
Suppose a point charge $Q$ is placed at the origin $O$ in free space. If a test charge $q$ is placed at a distance $r$ at point $P$ (where $OP = r$),then according to Coulomb's law,the force acting on $q$ is:
$\overrightarrow{F} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q q}{r^{2}} \hat{r}$
The electric field $\overrightarrow{E}$ at a point is defined as the force experienced by a unit positive test charge placed at that point:
$\overrightarrow{E} = \frac{\overrightarrow{F}}{q}$
Substituting the expression for force:
$\overrightarrow{E} = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{r^{2}} \hat{r}$ or $E = \frac{k Q}{r^{2}}$
Key Properties:
$1$. Electric field $\overrightarrow{E}$ is also known as electric field intensity.
$2$. The force acting on a charge $q$ at position vector $\vec{r}$ is given by $\overrightarrow{F}(\vec{r}) = q \overrightarrow{E}(\vec{r})$.
$3$. The $SI$ unit of electric field intensity is $N C^{-1}$ or $V m^{-1}$.
$4$. The dimensional formula is $[M^{1} L^{1} T^{-3} A^{-1}]$.
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View SolutionSix charges are placed around a regular hexagon of side length $a$ as shown in the figure. Five of them have charge $q$,and the remaining one has charge $x$. The perpendicular from each charge to the nearest hexagon side passes through the center $O$ of the hexagon and is bisected by the side.
Which of the following statement$(s)$ is(are) correct in $SI$ units?
$(A)$ When $x=q$,the magnitude of the electric field at $O$ is zero.
$(B)$ When $x=-q$,the magnitude of the electric field at $O$ is $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2q$,the potential at $O$ is $\frac{7q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3q$,the potential at $O$ is $\frac{3q}{4 \sqrt{3} \pi \epsilon_0 a}$.
Which of the following statement$(s)$ is(are) correct in $SI$ units?
$(A)$ When $x=q$,the magnitude of the electric field at $O$ is zero.
$(B)$ When $x=-q$,the magnitude of the electric field at $O$ is $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2q$,the potential at $O$ is $\frac{7q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3q$,the potential at $O$ is $\frac{3q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$A$ cube of side $L$ has point charges $+q$ located at its seven vertices and $-q$ at remaining one vertex. The electric field at its centre is found to be $|E|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is
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