Question

5 equal charges $‘q^{\prime }$ each are placed at the corners of a regular hexagon $\text{ABCDEF}$ of side $‘a^{\prime }$ as shown. Find the net electric field at the centre $\text{O}$ of the hexagon?

• A. $\frac{5kq}{a^2}$ towards $\text{F}$
• B. $\frac{kq}{a^2}$ towards $\text{F}$
• C. $3\sqrt{2}\frac{kq}{a^2}$ towards $\text{F}$
• D. $5\sqrt{2}\frac{kq}{a^2}$ towards $\text{C}$

Answer 1

The correct option is B $\frac{kq}{a^2}$ towards $\text{F}$

Inorder to solve, find electric field due to each charge at $\text{O}$ and then add using principle of superposition.

Distance between $\text{O}$ and $\text{B}$ is $‘a^{\prime }$ as side length of regular hexagon is given to be $‘a^{\prime }$.

Now ${\overrightarrow{E}}_{OB}=\frac{kq}{a^2}$ , directed radially away from $\text{B}$.

${\overrightarrow{E}}_{OE}=\frac{kq}{a^2}$; same magnitude but direction exactly opposite to ${\overrightarrow{E}}_{OB}$.

${\overrightarrow{E}}_{OE}$ and ${\overrightarrow{E}}_{OB}$ are equal and opposite vectors and hence will cancel each other out.

Similarly, ${\overrightarrow{E}}_{OA}$ and ${\overrightarrow{E}}_{OD}$ are equal and opposite, so they will also have a zero net effect.

Hence, net field will be only due to point charge at $\text{C}$.

${\overrightarrow{E}}_{OC}=\frac{kq}{a^2}$ direction away from $\text{C}$ i.e. towards $\text{F}$