What is the direction of the electric field at the centre O of the square in the figure shown below? Given that, q=10 nC and the side of the square is 5 cm .

Step-by-step solution

A D=B C= (5)^2+(5)^2 = 25+25 = 2 5 ~cm A O=B O=C O=O D= 5 2 2 ~cm The electric field, E= 1 4 ₀ q r^2 So, E_A= 9 10^9 10 10⁻⁹ 4 25 2 aligned & =7.2 NC ⁻¹ along O D E_B & = 9 10^9 2 10 10⁻⁹ 4 25 2 & =14.4 NC ⁻¹ along O B E_C & = 9 10^9 10 10⁻⁹ 4 25 2 & =7.2 along O C E_D & = 9 10^9 2 10 10⁻⁹ 4 25 2 & =14.4 along O A aligned Resultant of aligned E_A and E_D, E₁ & =(14.4-7.2) & =7.2 NC ⁻¹ along O A . aligned Resultant of E_B and E_C, E₂=(14.4-7.2)=7.2 NC ⁻¹ along O B . Since, E₁ and E₂ are perpendicular to each other.