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Chapter 17: Problem 105

A point charge \(q=-2.5 \mathrm{nC}\) is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is $4.0 \mu \mathrm{C} / \mathrm{m}^{2}\( and the space between the plates is \)6.0 \mathrm{mm} .$ (a) What is the potential difference between the plates? (b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?

Short Answer

Expert verified
Question: Calculate the potential difference between the plates and the kinetic energy of the point charge just before it hits the positive plate. Answer: To calculate the potential difference between the plates, use the formula: \(V = E \cdot d\) where \(E\) is the electric field calculated in step 1 and \(d\) is the distance between the plates (given as \(6.0 \mathrm{mm}\)). To calculate the kinetic energy of the point charge just before it hits the positive plate, apply energy conservation principle. The initial potential energy (\(U_i\)) will be equal to the final kinetic energy (\(K_f\)). Use the expression: \(K_f = (-2.5 × 10^{-9} \mathrm{C}) V\)

Step by step solution

01

Calculate the electric field between the plates

Given that the charge per unit area on the plates is \(4.0 \mu \mathrm{C} / \mathrm{m}^{2}\), we can calculate the electric field between the plates of the capacitor. The electric field (\(E\)) in a parallel-plate capacitor is related to the charge density (\(\sigma\)) as: \(E = \dfrac{\sigma}{\epsilon_0}\) where \(\epsilon_0 = 8.85 × 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)\) is the vacuum permittivity. By substituting the given values, we have: \(E = \dfrac{4.0 × 10^{-6} \mathrm{C/m}^2}{8.85 × 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)}\)
02

Calculate the potential difference between the plates

Using the electric field calculated in step 1, we can now calculate the potential difference (\(V\)) between the plates using the formula: \(V = E \cdot d\) where \(d\) is the distance between the plates. We are told that the distance is \(6.0 \mathrm{mm}\). Thus, we have: \(V = E \cdot 6.0 × 10^{-3} \mathrm{m}\)
03

Compute the initial potential energy of the point charge

We will now determine the initial potential energy of the point charge (\(U_i\)) placed adjacent to the negative plate using its charge (\(q\)) and the potential difference between the plates (\(V\)). This is calculated as: \(U_i = qV\) We are given the point charge \(q = -2.5 \mathrm{nC}\). Substitute the values into the equation: \(U_i = (-2.5 × 10^{-9} \mathrm{C}) V\)
04

Find the kinetic energy of the point charge

We are told that there are no other forces acting on the point charge, so we can use the energy conservation principle. The total initial energy (\(U_i\)) will be equal to the final kinetic energy (\(K_f\)) of the point charge just before it hits the positive plate. Thus: \(K_f = U_i\) Now, we can substitute the values and expressions from the previous steps to solve for \(K_f\). Then, the kinetic energy of the point charge just before it hits the positive plate is: \(K_f = (-2.5 × 10^{-9} \mathrm{C}) V\) Now, you can use these steps to calculate the potential difference between the plates and the kinetic energy of the point charge just before it hits the positive plate.

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Most popular questions from this chapter

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.]
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
An array of four charges is arranged along the \(x\) -axis at intervals of 1.0 \(\mathrm{m}\). (a) If two of the charges are \(+1.0 \mu \mathrm{C}\) and two are \(-1.0 \mu \mathrm{C},\) draw a configuration of these charges that minimizes the potential at \(x=0 .\) (b) If three of the charges are the same, $q=+1.0 \mu \mathrm{C},\( and the charge at the far right is \)-1.0 \mu \mathrm{C},$ what is the potential at the origin?
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