# Find the maximum KE of a charged Disc

• Hamiltonian
In summary, the conversation discusses how to find the initial potential energy and maximum kinetic energy of a system consisting of a disc and a hollow sphere. By assuming the sphere to be made up of two hemispheres, the initial potential energy is calculated to be ##U_0 = \frac{\sigma^2 \pi R^3}{2\epsilon_0}##. The maximum kinetic energy is then found by conserving energy and linear momentum, resulting in the equation ##{KE}_{max} = \frac{2}{5}\frac{\sigma^2 \pi R^3}{\epsilon_0}##. The conversation also mentions some potential errors in the calculations and suggests using the relationship ##KE = \frac{p
Hamiltonian
Homework Statement
A thin disc of Radius R is held closing the opening of a thin hemispherical shell of the same radius. Both the bodies are made of insulating materials and have a uniform surface charge density ##\sigma## and surface mass density. When the system is released what will be the maximum KE of the disc?
Relevant Equations
##V = \frac{K q_1 q_2}{r}##, Conservation of energy&linear momentum
To find the initial potential energy of the system we can assume the disc to be placed inside a hollow sphere of the same radius and ##\sigma##, the potential energy inside a charged hollow shell is:
$$V = \frac{\sigma(4\pi R^2)}{4\pi \epsilon_0 R} = \frac{\sigma R}{\epsilon_0}$$
the potential energy of the sphere and disc system will be:
$$U = V(\sigma \pi R^2) = \frac{\sigma^2\pi R^3}{\epsilon_0}$$
##U## is the potential energy due to the disc and the sphere, but we need to find the potential energy of the disc and hemisphere system, the sphere can be assumed to be made up of two hemispheres hence each contributing to ##U/2##
therefore the initial potential energy is:
$$U_0 = \frac{\sigma^2 \pi R^3}{2\epsilon_0}$$
the disc will have maximum KE when it's at ##\infty## so the potential energy of the system ##\rightarrow 0##
(assume surface mass density of both hemisphere and disc be ##\lambda##)
hence we can conserve energy and linear momentum and solve for the maximum KE of the disc
$$\frac{\sigma^2 \pi R^3}{2\epsilon} = \frac{1}{2}(\lambda \pi R^2){v_d}^2 + \frac{1}{2}(\lambda 4\pi R^2){v_h}^2$$
$$\lambda \pi R^2 v_d = \lambda 4 \pi R^2 v_h \implies v_h = \frac{v_d}{4}$$
where ##v_h## is the velocity of the hemisphere and ##v_d## is the velocity of the disc at ##\infty##
solving the above equations gives:
$${v_d}^2 = \frac{4}{5}\frac{\sigma^2 R}{\epsilon_0 \lambda} \implies{KE}_{max} = \frac{2}{5}\frac{\sigma^2 \pi R^3}{\epsilon_0}$$

I don't have the correct answer to this question so can someone check my work and the final answer?

Hamiltonian299792458 said:
hence we can conserve energy and linear momentum and solve for the maximum KE of the disc
$$\frac{\sigma^2 \pi R^3}{2\epsilon} = \frac{1}{2}(\lambda \pi R^2){v_d}^2 + \frac{1}{2}(\lambda 4\pi R^2){v_h}^2$$
$$\lambda \pi R^2 v_d = \lambda 4 \pi R^2 v_h \implies v_h = \frac{v_d}{4}$$
where ##v_h## is the velocity of the hemisphere and ##v_d## is the velocity of the disc at ##\infty##
solving the above equations gives:
$${v_d}^2 = \frac{4}{5}\frac{\sigma^2 R}{\epsilon_0 \lambda} \implies{KE}_{max} = \frac{2}{5}\frac{\sigma^2 \pi R^3}{\epsilon_0}$$
It looks like you have considered the disc and an entire spherical shell moving away from each other.

The disc’s area is ##\pi R^2## and the hemisphere’s is ##2\pi R^2## (not ##4\pi R^2##). I.e. the mass of the disc is half the mass of the hemisphere. Conservation of momentum then immediately tells you that ##v_h= \frac {v_d}{2}## in the centre-of-mass frame.

To get the split in kinetic energy you might consider using the handy relationship ##KE = \frac{p^2}{2m}## and using simple proportionality.

Hamiltonian

## What is the formula for finding the maximum KE of a charged disc?

The formula for finding the maximum KE of a charged disc is KE = (1/2)mv^2, where m is the mass of the disc and v is the velocity at which it is moving.

## How do you calculate the mass of a charged disc?

The mass of a charged disc can be calculated by using a balance scale or by measuring its weight with a calibrated scale. Alternatively, the mass can be calculated by dividing the disc's density by its volume.

## Can the maximum KE of a charged disc be negative?

No, the maximum KE of a charged disc cannot be negative. KE is a measure of energy, and energy cannot be negative. However, the velocity of the disc can be negative, which would result in a negative value for KE. In this case, the disc is moving in the opposite direction of the chosen reference point.

## How does the charge of the disc affect its maximum KE?

The charge of the disc does not directly affect its maximum KE. However, a charged disc may experience a force due to its interaction with other charged objects, which can change its velocity and therefore its KE.

## What factors can affect the maximum KE of a charged disc?

The maximum KE of a charged disc can be affected by its mass, velocity, and any external forces acting on it. Additionally, the presence of other charged objects in the vicinity can also affect the disc's KE due to electromagnetic interactions.

• Introductory Physics Homework Help
Replies
12
Views
198
• Introductory Physics Homework Help
Replies
23
Views
405
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
916
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
138
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
513
• Introductory Physics Homework Help
Replies
11
Views
740