Question
State and explain principle of superposition.
✓
Answer
Statement: When a number of charges are interacting, the resultant force on a particular charge is given by the vector sum of the forces exerted by individual charges.
Explanation:
i. Consider a number of point charges q1, q2, q3 ……………… kept at points A1, A2, A3 ………….. as shown in figure
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ii. The force exerted on the charge $q _1$ by $q _2$ is $\vec{F}_{12}$ The value of $\vec{F}_{12}$ is calculated by ignoring the presence of other charges. Similarly, force $\vec{F}_{13}, \vec{F}_{14}$ can be found, using the Coulomb's law.
iii. Total force $\vec{F}_1$ on charge qi is the vector sum of all such forces.
$\vec{F}_1=\vec{F}_{12}+\vec{F}_{13}+\vec{F}_{14}+$ .............
$=\frac{1}{4 \pi \varepsilon_0}\left[\frac{ q _1 q _2}{\left| r _{21}\right|^2} \times \hat{ r }_{21}+\frac{ q _1 q _3}{\left| r _{31}\right|^2} \times \hat{ r }_{31}+\ldots\right]$
where $\hat{r}_{21}, \hat{r}_{31}$ are unit vectors directed to $q _1$ from $q _2, q_3$ respectively and $r_{21}, r_{31}, r_{41}$ are the distances from $q_1$ to $q_2, q_3$ respectively.
iv. If $q_1, q_2, q_3 \ldots . . . ., q_n$ are the point charges then the force $\vec{F}$ exerted by these charges on a test charge $q _0$ is given by,
$\vec{F}_{\text {test }}=\vec{F}_1=\vec{F}_2+\vec{F}_3+\ldots+\vec{F}_{ n }$
$=\sum_{ n =1}^{ n } F _{ n }=\frac{1}{4 \pi \varepsilon_0} \sum_{ n =1}^{ n } \frac{ q _0 q _{ n }}{ r _{ n }^2} \hat{ r }_{ n }$
Where, $\hat{r}_n$ is a unit vector directed from the nth charge to the test charge $q_0$ and $r_2$ is the separation between them, $\vec{r}_{ n }= r _{ n } \hat{r}_{ n }$
Explanation:
i. Consider a number of point charges q1, q2, q3 ……………… kept at points A1, A2, A3 ………….. as shown in figure
ii. The force exerted on the charge $q _1$ by $q _2$ is $\vec{F}_{12}$ The value of $\vec{F}_{12}$ is calculated by ignoring the presence of other charges. Similarly, force $\vec{F}_{13}, \vec{F}_{14}$ can be found, using the Coulomb's law.
iii. Total force $\vec{F}_1$ on charge qi is the vector sum of all such forces.
$\vec{F}_1=\vec{F}_{12}+\vec{F}_{13}+\vec{F}_{14}+$ .............
$=\frac{1}{4 \pi \varepsilon_0}\left[\frac{ q _1 q _2}{\left| r _{21}\right|^2} \times \hat{ r }_{21}+\frac{ q _1 q _3}{\left| r _{31}\right|^2} \times \hat{ r }_{31}+\ldots\right]$
where $\hat{r}_{21}, \hat{r}_{31}$ are unit vectors directed to $q _1$ from $q _2, q_3$ respectively and $r_{21}, r_{31}, r_{41}$ are the distances from $q_1$ to $q_2, q_3$ respectively.
iv. If $q_1, q_2, q_3 \ldots . . . ., q_n$ are the point charges then the force $\vec{F}$ exerted by these charges on a test charge $q _0$ is given by,
$\vec{F}_{\text {test }}=\vec{F}_1=\vec{F}_2+\vec{F}_3+\ldots+\vec{F}_{ n }$
$=\sum_{ n =1}^{ n } F _{ n }=\frac{1}{4 \pi \varepsilon_0} \sum_{ n =1}^{ n } \frac{ q _0 q _{ n }}{ r _{ n }^2} \hat{ r }_{ n }$
Where, $\hat{r}_n$ is a unit vector directed from the nth charge to the test charge $q_0$ and $r_2$ is the separation between them, $\vec{r}_{ n }= r _{ n } \hat{r}_{ n }$
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