Question 169: Electrostatics & Capacitance

Question

Five positive charges each having charge $q$ are placed at the vertices of a pentagon as shown in the figure. The electric potential $(V)$ and the electric field $(\vec{E})$ at the center $O$ of the pentagon due to these five positive charges are :

Accepted Answer

Electric potential is a scalar quantity. According to the principle of superposition the total electric potential at a point is the algebraic sum of the individual potentials created by each charge.
The potential $\mathrm{V}_{\mathrm{i}}$ at the center (at distance r ) due to a single point charge +q at a vertex is :
$$ \mathrm{V}_{\mathrm{i}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}} $$
Since there are 5 identical positive charges situated at the exact same distance r from the center O , the total potential V is simply 5 times the potential of a single charge :
$$ V=V_1+V_2+V_3+V_4+V_5 $$
$$\Rightarrow $$ $\mathrm{V}=5\left(\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}}ight)$
$$ \mathrm{V}=\frac{\mathrm{5q}}{4 \pi \epsilon_0 \mathrm{r}} $$
Electric field is a vector quantity, which means both magnitude and direction must be considered.
The net electric field at the centre is the vector sum of the electric fields produced by each individual charge :
$$ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=\overrightarrow{\mathrm{E}}_1+\overrightarrow{\mathrm{E}}_2+\overrightarrow{\mathrm{E}}_3+\overrightarrow{\mathrm{E}}_4+\overrightarrow{\mathrm{E}}_5 $$
$$\Rightarrow $$ $\overrightarrow{\mathrm{E}}_{\mathrm{net}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_1+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_2+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_3+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_4+\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} \hat{\mathrm{r}}_5$
$$\Rightarrow $$ $\overrightarrow{\mathrm{E}}_1=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\left(\hat{r}_1+\hat{r}_2+\hat{r}_3+\hat{r}_4+\hat{r}_5ight)$
$$\Rightarrow $$ $$ \overrightarrow{\mathrm{E}}_{\mathrm{net}}=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\left(\frac{\overrightarrow{\mathrm{r}}_1}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_2}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_3}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_4}{\mathrm{r}}+\frac{\overrightarrow{\mathrm{r}}_5}{\mathrm{r}}ight)=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}}{\mathrm{r}^3}\left(\overrightarrow{\mathrm{r}}_1+\overrightarrow{\mathrm{r}}_2+\overrightarrow{\mathrm{r}}_3+\overrightarrow{\mathrm{r}}_4+\overrightarrow{\mathrm{r}}_5ight) $$
Using closed polygon rule, $\vec{r}_1+\vec{r}_2+\vec{r}_3+\vec{r}_4+\vec{r}_5=\overrightarrow{0}$
Therefore, $\overrightarrow{\mathrm{E}}_{	ext {net }}=0$
So, at the centre of regular polygon,
The electric potential is $V=\frac{5 q}{4 \pi \epsilon_0 r}$ and the electric field is $\vec{E}=0$.
Hence, the correct option is (B).

Answer

$V=\frac{5 q}{4 \pi \varepsilon_o r}$ and $\vec{E}=\frac{5 \sqrt{3} q}{8 \pi \varepsilon_o r^2} \hat{r}$

Answer

$V=\frac{5 q}{4 \pi \varepsilon_o r}$ and $\vec{E}=0$

Answer

$V=0$ and $\vec{E}=0$

Answer

$V=\frac{5 q}{4 \pi \varepsilon_0 r}$ and $\vec{E}=\frac{5 q}{4 \pi \varepsilon_{\mathrm{o}} r^2} \hat{r}$

Electrostatics & Capacitance

Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field (\vec{E}) at the center O of the pentagon due to these five positive charges are :

Choose the correct answer: