Question

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

Find E due to each charge then add by principle of superposition.

Distance between O and B is ‘a’ as side length of regular hexagon is given to be ‘a’

Now ${\overrightarrow{E}}_{OB}=\frac{kq}{a^2}$ , directedradially away from 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.

Net field will be only due to point charge at C

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