Question

Two pellets, each with a charge of 0.70 microcoulomb (7.0×10^-7 C ), are located 4.0 cm (4.0×10^-2 m ) apart. Find the electric force between them.What would be the mass of an object that would experience this same force in Earth's gravitational field?

Answer 1

`0.2812 " kg"`

Explanation:

Given:

• Charge on each pellet(q) `= 7 xx 10^(-7) " C"`
• Distance between pellet(r) `= 4 xx 10^(-2) " m"`
• `K= 9 xx 10^9 (" N""m"^2)/("c"^2)`

So electrostatic force between them:

#F= Kq^2/r^2 = (9 xx 10^9) (7 xx 10^(-7))^2/(4 xx 10^(-2))^2 = 2.756 " N"#

Now force of gravitation:

`F_g= mg = 2.756 " N"`

Mass of object that would experience the same force

`m= 2.756/9.8= 0.2812 " kg"`