To determine the final charge on each sphere, follow these steps:
Step 1: Initial Conditions
• Two metallic spheres are initially at a distance r=0.50 mr = 0.50 \, \text{m} apart, having charges q1q_1 and q2q_2, with q1≠−q2q_1 \neq -q_2 (unequal and opposite charges).
• After contact, the charges redistribute equally because the spheres are identical. Let the final charge on each sphere be qq.
Step 2: Coulomb's Force After Contact
After contact, the spheres are separated again to the same distance r=0.50 mr = 0.50 \, \text{m}, and the force of repulsion between them is F=0.108 NF = 0.108 \, \text{N}.
Using Coulomb's Law:
F=kq2r2,F = \frac{k q^2}{r^2},
where:
• F=0.108 NF = 0.108 \, \text{N},
• k=9×109 N\cdotpm2/C2k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 (Coulomb constant),
• r=0.50 mr = 0.50 \, \text{m},
• qq is the final charge on each sphere.
Rearranging for qq:
q2=Fr2k.q^2 = \frac{F r^2}{k}.
Substitute the values:
q2=(0.108)(0.50)29×109.q^2 = \frac{(0.108)(0.50)^2}{9 \times 10^9}.
Simplify:
q2=(0.108)(0.25)9×109,q^2 = \frac{(0.108)(0.25)}{9 \times 10^9}, q2=0.0279×109,q^2 = \frac{0.027}{9 \times 10^9}, q2=3×10−12.q^2 = 3 \times 10^{-12}.
Taking the square root:
q=3×10−12=1.73×10−6 C.q = \sqrt{3 \times 10^{-12}} = 1.73 \times 10^{-6} \, \text{C}.
Step 3: Final Charge on Each Sphere
After contact, the final charge on each sphere is:
q=1.73 μC.q = 1.73 \, \mu\text{C}.
Step 4: Conclusion
The final charge on each sphere after contact is q=1.73 μCq = 1.73 \, \mu\text{C}.
