Two identical isolated metal spheres carrying charges of +8.0 \times 10^{-6}\text{ C} and … — JAMB UTME

Two identical isolated metal spheres carrying charges of $+8.0 \times 10^{-6}\text{ C}$ and $-2.0 \times 10^{-6}\text{ C}$ are brought into contact and then separated to a distance of $0.30\text{ m}$ in a vacuum. What is the magnitude of the electrostatic force of repulsion, in newtons ( $\text{N}$ ), between the spheres after contact? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{C}^{-2}$ )

Cevap: 0.9 N

Cevap

The magnitude of the electrostatic force of repulsion between the spheres after contact is

0.9\text{ N}

When identical conducting spheres touch, their total electric charge is conserved and shared equally. The net charge is

+8.0\,\mu\text{C} + (-2.0\,\mu\text{C}) = +6.0\,\mu\text{C}

, giving each sphere a charge of

+3.0\,\mu\text{C}

. Applying Coulomb's law with a distance of

0.30\text{ m}

yields

0.9\text{ N}

Adım Adım Çözüm

Calculate the net combined charge of the two identical spheres when brought into contact.

Q_{\text{total}} = q_1 + q_2 = (+8.0 \times 10^{-6}\text{ C}) + (-2.0 \times 10^{-6}\text{ C}) = +6.0 \times 10^{-6}\text{ C}

According to the principle of conservation of charge, charges add algebraically.

Determine the charge on each individual sphere after separation.

q' = \frac{Q_{\text{total}}}{2} = \frac{+6.0 \times 10^{-6}\text{ C}}{2} = +3.0 \times 10^{-6}\text{ C}

Identical conducting spheres share total charge equally when in contact.

Compute the force of repulsion using Coulomb's law.

F = \frac{k(q')^2}{r^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})^2}{(0.30)^2} = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-12}}{0.09} = 0.9\text{ N}

Coulomb's law defines the electrostatic force between two point charges.

Anahtar Kavram

Charge conservation, redistribution by conduction, and Coulomb's law

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