Question 25: Electrostats & 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 $V_i$ at the center (at distance $r$) due to a single point charge $+q$ 
$at a vertex is: $$Vi = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$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:$$V = V_1 + V_2 + V_3 + V_4 + V_5$$$$\Rightarrow V = 5 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r} ight)$$$$V = \frac{5q}{4\pi\epsilon_0 r}$$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:$$\vec{E}_{	ext{net}} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4 + \vec{E}_5$$$$\Rightarrow \vec{E}_{	ext{net}} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}_1 + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}_2 + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}_3 + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}_4 + \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r}_5$$$$\Rightarrow \vec{E}_{	ext{net}} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} (\hat{r}_1 + \hat{r}_2 + \hat{r}_3 + \hat{r}_4 + \hat{r}_5)$$$$\Rightarrow \vec{E}_{	ext{net}} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \left( \frac{\vec{r}_1}{r} + \frac{\vec{r}_2}{r} + \frac{\vec{r}_3}{r} + \frac{\vec{r}_4}{r} + \frac{\vec{r}_5}{r} ight) = \frac{1}{4\pi\epsilon_0} \frac{q}{r^3} (\vec{r}_1 + \vec{r}_2 + \vec{r}_3 + \vec{r}_4 + \vec{r}_5)$$Using closed polygon rule, $\vec{r}_1 + \vec{r}_2 + \vec{r}_3 + \vec{r}_4 + \vec{r}_5 = \vec{0}$Therefore, $\vec{E}_{	ext{net}} = 0$So, at the centre of regular polygon,The electric potential is $V = \frac{5q}{4\pi\epsilon_0 r}$ and the electric field is $\$\vec{E} = 0$.

Answer

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

Answer

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

Answer

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

Answer

D. $V = \frac{5q}{4\pi\epsilon_o r}$ and $\vec{E} = \frac{5q}{4\pi\epsilon_o r^2} \hat{r}$

Electrostats & 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 :

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