Question

regular hexagon with side length a is shown in figure find e​

Answer 1

Answer: Electric field at A is 3kQ/2a²

Explanation:

Electric field at A will be vector sum of electric field due to all charges.

Electric field due to 2Q charge = $\frac{k2Q}{(2a)^{2} }$

Electric field due to 2Q charge = kQ/2a² towards left

Electric field due to each Q charge = $\frac{kQ}{a^{2} }$

Vertical component of the field of Q charge cancel each other.

Horizontal component gets added up.

Horizontal component due to each Q charge = $\frac{kQ}{a^{2} }cos(60)$ = kQ/2a²

Total horizontal field due to Q charges = kQ/a² towards left

Net electric field =   $\frac{kQ}{2a^{2} } + \frac{kQ}{a^{2} }$

Net electric field = $\frac{3kQ}{2a^{2} }$

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