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Electric Charges and Fields — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields5 Marks
Question
Write the definition of electrostatic at any point. Also write its S.I. units. As shown in the figure, three points charges $q_1$, $q _2$ and $q _3$ are placed at points $A, B$ and $C$ respectively. Drive an expression for the electrostatic potential energy of this system.

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Answer

Electrostatic Potential : "Electric potential at any point inside the electric field is equal to the ratio of the work W done in bringing a test positive charge from infinity to that point without acceleration and the value of the test charge $q_0$." The unit of electric potential in SI system is Volt. If charge $q_1$ is brought from infinity to $\vec{r}_1$, no work is done. Therefore $W _1=0$.
Work done in bringing charge $q _2$ from infinity to $\overrightarrow{r_2}$,
$ W_{12}=q_2, V_1=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{12}} $
where the potential at distance $r_{12}$ is $V _1$.
Now charges $q_1$ and $q_2$ produce potential at the distance $\vec{r}_3$
$ V_3=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{13}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_{23}} $
Work done in bringing the charge $q_{13}$ from infinity to $\overrightarrow{r_3}$,
$W ^{\prime}= W _{13}+ W _{23}=q_3 V_3$
$=q_3\left[\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_{13}}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_{23}}\right]$
$=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{r_{13}}+\frac{1}{4 \pi \varepsilon_0} \frac{q_2 q_3}{r_{23}}$
Therefore, the value of electrostatic potential energy of this system is
$U = W _{12}+ W ^{\prime}$
$=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r_{12}}+\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_3}{r_{13}}+\frac{1}{4 \pi \varepsilon_0} \frac{q_2 q_3}{r_{23}}$
$=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r_{12}}+\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right)$

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