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Alenkinab [10]
3 years ago
13

Find net force acting on O

Physics
1 answer:
klemol [59]3 years ago
8 0

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

<u>Explanation:</u>

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:

F = k({\frac{Qq_1}{r^2} + \frac{Qq_2}{r^2} + \frac{Qq_3}{r^2} + \frac{Qq_4}{r^2} )\\\\\\

F = \frac{kQ}{r^2} (q_1 + q_2 + q_3 + q_4)

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

k = 8.99 x 10⁹ N m2 / C2

On substituting the value, we get:

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

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

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