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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: \(K_f = (-2.5 脳 10^{-9} \mathrm{C}) V\)

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 \(\epsilon_0 = 8.85 脳 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)\) is the vacuum permittivity. By substituting the given values, we have: \(E = \dfrac{4.0 脳 10^{-6} \mathrm{C/m}^2}{8.85 脳 10^{-12} \mathrm{C}^2/(\mathrm{N} \cdot \mathrm{m}^2)}\)
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: \(V = E \cdot 6.0 脳 10^{-3} \mathrm{m}\)
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: \(U_i = (-2.5 脳 10^{-9} \mathrm{C}) V\)
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: \(K_f = (-2.5 脳 10^{-9} \mathrm{C}) V\) 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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