The net force acting on the center O is 4.495 X 10⁹ N
Explanation:
Given:
Charge-
q₁ = 4C
q₂ = 3C
q₃ = -5C
q₄ = 2C
q₅ = 1C
Side of the square = 4cm
Net force acting on the center, F = ?
Length of the side of the square is 4cm
The length of the diagonal will be 4√2
The length of half diagonal = 4√2 / 2 = 2√2
Force acting on the center of the square is:
[tex]F = k({\frac{Qq_1}{r^2} + \frac{Qq_2}{r^2} + \frac{Qq_3}{r^2} + \frac{Qq_4}{r^2} )\\\\\\[/tex]
[tex]F = \frac{kQ}{r^2} (q_1 + q_2 + q_3 + q_4)[/tex]
Here, k is coulomb's constant and the value is
k = 8.99 x 10⁹ N m2 / C2
On substituting the value, we get:
[tex]F = \frac{8.99 X 10^9 X 1}{(2\sqrt{2} )^2} ( 4C + 3C - 5C + 2C)\\\\F = \frac{8.99 X 10^9}{8} (4C)\\\\F = 4.495 X 10^9 N[/tex]
Therefore, the net force acting on the center O is 4.495 X 10⁹ N
