Question

A metallic sphere has a charge of \(+4.0 \mathrm{nC} .\) A negatively charged rod has a charge of \(-6.0 \mathrm{nC} .\) When the rod touches the sphere, $8.2 \times 10^{9}$ electrons are transferred. What are the charges of the sphere and the rod now?

Short answer

The new charge of the sphere is \(2.688 \times 10^{-9} C\), and the new charge of the rod is \(-4.688 \times 10^{-9} C\).

Step-by-step solution

Calculate the charge of one electron

The charge of one electron (e) is approximately -1.6 x 10^{-19} C. **Step 2: Calculate the Charge Transfer**

Find the total charge transferred

The number of electrons transferred is \(8.2 \times 10^9\). We can calculate the charge transfer (q) by multiplying the charge of one electron and the number of electrons: q = (8.2 x 10^9) x (-1.6 x 10^{-19} C) = -1.312 x 10^{-9} C **Step 3: Calculate the New Charge of the Sphere**

Find the new charge of the sphere

The initial charge of the sphere (q1) is +4.0 nC. The sphere will receive a negative charge of -1.312 x 10^{-9} C. We can calculate the final charge (q1') by adding the charge transfer to the initial charge: q1' = q1 + q = (+4.0 x 10^{-9} C) + (-1.312 x 10^{-9} C) = 2.688 x 10^{-9} C **Step 4: Calculate the New Charge of the Rod**

Find the new charge of the rod

The initial charge of the rod (q2) is -6.0 nC. The rod will lose the same charge (-1.312 x 10^{-9} C) transferred to the sphere. We can calculate the new charge (q2') by adding the charge transfer to the initial charge: q2' = q2 - q = (-6.0 x 10^{-9} C) - (-1.312 x 10^{-9} C) = -4.688 x 10^{-9} C **Step 5: Result**

State the final charges of the sphere and the rod

The final charge of the sphere is \(q1' = 2.688 \times 10^{-9} C\), and the final charge of the rod is \(q2' = -4.688 \times 10^{-9} C\).