Find net force acting on O

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:

Here, k is coulomb's constant and the value is

k = 8.99 x 10⁹ N m2 / C2

On substituting the value, we get:

Therefore, the net force acting on the center O is 4.495 X 10⁹ N

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg