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Chapter 17: Problem 88

Charges of \(-12.0 \mathrm{nC}\) and \(-22.0 \mathrm{nC}\) are separated by $0.700 \mathrm{m} .$ What is the potential midway between the two charges?

Short Answer

Expert verified
Question: Determine the electric potential at the midpoint between two charges of \(-12.0 \, \mathrm{nC}\) and \(-22.0 \, \mathrm{nC}\) separated by a distance of \(0.700 \, \mathrm{m}\). Answer: The electric potential at the midpoint between the two charges is \(-876 \, \mathrm{V}\).

Step by step solution

01

Convert the charges to coulombs

We need to convert the charges from nanocoulombs to coulombs. Recall that 1 nC = \(10^{-9}\) C. \(-12.0 \, \mathrm{nC} = -12.0 \times 10^{-9} \, \mathrm{C}\) \(-22.0 \, \mathrm{nC} = -22.0 \times 10^{-9} \, \mathrm{C}\)
02

Find the distance to the midpoint

The distance between the charges is \(0.700 \, \mathrm{m}\), so the distance to the midpoint is half of that: \(r_{mid} = \frac{0.700}{2} \, \mathrm{m} = 0.350 \, \mathrm{m}\)
03

Find the electric potential due to each charge at the midpoint

We can use the formula \(V = k\frac{q}{r}\), where \(k\) is the electrostatic constant (\(k = 8.99 \times 10^9 \, \mathrm{Nm^2/C^2}\)). For the \(-12.0 \times 10^{-9} \, \mathrm{C}\) charge: \(V_1 = k\frac{q_1}{r_{mid}} = (8.99 \times 10^9 \, \mathrm{Nm^2/C^2})(-12.0 \times 10^{-9} \, \mathrm{C}) / (0.350 \, \mathrm{m})\) For the \(-22.0 \times 10^{-9} \, \mathrm{C}\) charge: \(V_2 = k\frac{q_2}{r_{mid}} = (8.99 \times 10^9 \, \mathrm{Nm^2/C^2})(-22.0 \times 10^{-9} \, \mathrm{C}) / (0.350 \, \mathrm{m})\)
04

Sum the electric potentials to find the total potential at the midpoint

We can simply add the potential due to each charge to find the total potential at the midpoint. \(V_{total} = V_1 + V_2\) Plug in the values we found in Step 3 and perform the calculation. \(V_{total} = (-3.08 \times 10^2 \, \mathrm{V}) + (-5.68 \times 10^2 \, \mathrm{V}) = -8.76 \times 10^2 \, \mathrm{V}\) So, the electric potential at the midpoint between the two charges is \(-876 \, \mathrm{V}\).

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Most popular questions from this chapter

A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
A tiny hole is made in the center of the negatively and positively charged plates of a capacitor, allowing a beam of electrons to pass through and emerge from the far side. If \(40.0 \mathrm{V}\) are applied across the capacitor plates and the electrons enter through the hole in the negatively charged plate with a speed of \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) what is the speed of the electrons as they emerge from the hole in the positive plate?
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of magnitude \(0.800 \mu\) C on each plate. (a) What is the potential difference between the plates? (b) If the plate separation is doubled, while the charge is kept constant, what will happen to the potential difference?
A spherical conductor of radius \(R\) carries a total charge Q. (a) Show that the magnitude of the electric field just outside the sphere is \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the charge per unit area on the conductor's surface. (b) Construct an argument to show why the electric field at a point \(P\) just outside any conductor in electrostatic equilibrium has magnitude \(E=\sigma / \epsilon_{0},\) where \(\sigma\) is the local surface charge density. [Hint: Consider a tiny area of an arbitrary conductor and compare it to an area of the same size on a spherical conductor with the same charge density. Think about the number of field lines starting or ending on the two areas.]
By rewriting each unit in terms of kilograms, meters, seconds, and coulombs, show that \(1 \mathrm{N} / \mathrm{C}=1 \mathrm{V} / \mathrm{m}\)
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