Question 557 of 763 | CUET (Common University Entrance Test) PG Material Science (SCQP18)

As shown in figure, in a regular hexagon, ${\displaystyle ∆AOB}$ is an equilateral triangle. Therefore distance from vertex to centre of the hexagon is ${\displaystyle a}$.

Now, Electric potential is a scalar quantity. It is the amount of work needed to move a unit positive charge from a reference point to specific point inside the field without producing any acceleration.

The electric potential created by a point charge ${\displaystyle q}$ at a distance ${\displaystyle r}$ from the charge, is given as;

${\displaystyle V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}}$

Here, at centre of the hexagon, electrostatic potential is due to five equal charges. Therefore potential at centre ${\displaystyle O}$ is,

${\displaystyle V_0=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}}$

${\displaystyle \therefore V_0=\frac{1}{4\pi {\epsilon }_0}\frac{5q}{r}}$

The electric field is a vector quantity. The electric field produced due to a charge ${\displaystyle q}$, at distance ${\displaystyle r}$ from charge is given as,

${\displaystyle \stackrel{⇾}{E}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r}}$

Where, ${\displaystyle \widehat{r}}$ is a unit vector of electric field vector.

Here, all five charges are equal. And distances from centre to each vertices are also same ${\displaystyle a}$. Therefore, magnitude of each electric fields at the centre of hexagon are equal.

${\displaystyle \left|\stackrel{⇾}{E_A}\right|=\left|\stackrel{⇾}{E_B}\right|=\left|\stackrel{⇾}{E_D}\right|=\left|\stackrel{⇾}{E_E}\right|=\left|\stackrel{⇾}{E_F}\right|}$

As shown in figure, electric field due to charge at point ${\displaystyle A}$ is directed opposite to electric field due to charge at point ${\displaystyle D}$.

${\displaystyle \therefore \stackrel{⇾}{E_A}=-\stackrel{⇾}{E_D}}$

Therefore, effective electric field of charges at points ${\displaystyle A}$ and ${\displaystyle D}$ becomes zero.

Similarly, effective electric field due to charges at points ${\displaystyle E}$ and ${\displaystyle B}$ also becomes zero.

It means net electric field at the centre of the hexagon is only due to charge at point ${\displaystyle F}$. Therefore, magnitude of electric field at the centre of the hexagon is,

${\displaystyle E_O=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}}$