Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 49.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0705 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1 < q2. The Coulomb Force constant = 1/(4piEknot) =8.99x10^9 Nxm^2/C^2

q1= (blank)C

q2=(blank)C

Answer 1

r = distance between the spheres = 49 cm = 0.49 m

F_i = force between the spheres initially = 0.0705 N

electric force between the spheres is given as

F_i = k q₁ q₂/r²

0.0705 = (9 x 10⁹) q₁ q₂/(0.49)²

q₁ q₂ = 1.88 x 10^-12 eq-1

after the spheres are connected using conducting wire, the charge redistribute until they have equal charge. let the charge on each be "q" after disconnecting

new force is given as

F_n = k q²/r²

0.115 = (9 x 10⁹) q²/(0.49)²

q = 1.75 x 10^-6

using conservation of charge

q₁ + q₂ = 2q

q₁ + q₂ = 2 (1.75 x 10^-6)

q₁ + q₂ = 3.5 x 10^-6   

q₁ = (3.5 x 10^-6) - q₂ eq-2

using eq-1 and eq-2

((3.5 x 10^-6) - q₂) q₂ = 1.88 x 10^-12

q₂ = 2.84 x 10^-6 C

using eq-2

q₁ = (3.5 x 10^-6) - q₂  

q₁ = (3.5 x 10^-6) - ( 2.84 x 10^-6)  

q₁ = 6.625 x 10^-7 C