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Chapter 16: Problem 4

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

Expert verified
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

01

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**
02

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**
03

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**
04

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**
05

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 \(q2' = -4.688 \times 10^{-9} C\).

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Most popular questions from this chapter

A point charge \(q_{1}=+5.0 \mu \mathrm{C}\) is fixed in place at \(x=0\) and a point charge \(q_{2}=-3.0 \mu \mathrm{C}\) is fixed at \(x=\) $-20.0 \mathrm{cm} .\( Where can we place a point charge \)q_{3}=-8.0 \mu \mathrm{C}$ so that the net electric force on \(q_{1}\) due to \(q_{2}\) and \(q_{3}\) is zero?
What is the ratio of the electric force to the gravitational force between a proton and an electron separated by \(5.3 \times 10^{-11} \mathrm{~m}\) (the radius of a hydrogen atom)?
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$

An equilateral triangle has a point charge \(+q\) at each of the three vertices \((A, B, C) .\) Another point charge \(Q\) is placed at \(D\), the midpoint of the side \(B C\). Solve for \(Q\) if the total electric force on the charge at \(A\) due to the charges at \(B, C,\) and \(D\) is zero.
Two point charges are located on the \(x\) -axis: a charge of \(+6.0 \mathrm{nC}\) at \(x=0\) and an unknown charge \(q\) at \(x=\) \(0.50 \mathrm{m} .\) No other charges are nearby. If the electric field is zero at the point $x=1.0 \mathrm{m},\( what is \)q ?$
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