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Q.4) Two table tennis balls hang with their centers 10.0 cm apart. The charge on ball
A is +12x10°C, and the charge on Ball B is -15x10°C. What is the force of
attraction between the balls?


Sagot :

Answer:

F = 1.62 x 10¹⁴ N

Explanation:

The electrostatic force of attraction between the balls can be given by the following formula:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where,

F = attractive force = ?

k = Colomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Charge on ball A = 12 C

q₂ = Charge on ball B = 15 C

r = distanc between the balls = 10 cm = 0.1 m

Therefore,

[tex]F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(12\ C)(15\ C)}{(0.1\ m)^2}[/tex]

F = 1.62 x 10¹⁴ N