# Electrostatics problem involving infinite sheet

• Tanya Sharma
In summary, the problem involves an infinite dielectric sheet with a hole of radius R and charge density σ. An electron is released at a distance R√3 from the centre of the hole and the question asks for its velocity when it crosses the plane of the sheet. The solution involves finding the electric field at a distance x due to the sheet and the circular disk containing charge density -σ, and then using the equation mv(dv/dx) = -σx/(2ε0√(R^2+x^2)) to calculate the velocity. Integrating under proper limits, the final result is v = √(σeR/(mε0)). Alternatively, the problem can be approached using energy conservation,
Tanya Sharma

## Homework Statement

An infinite dielectric sheet having charge density σ has a hole of radius R in it. An electron is released on the axis of the hole at a distance R√3 from the centre. What will be the velocity which it crosses the plane of sheet? (e = charge on electron and m = mass of electron)

## The Attempt at a Solution

Force at a distance x on the axis of the hole can be calculated by finding electric field at a distance x .Electric field is found by superimposing the electric field due to the sheet and the circular disk containing charge density -σ .

## E = \frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##mv\frac{dv}{dx} = -\frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##vdv = -\frac{σ}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

Now integrating under proper limits I get ## v^2 = \frac{\sqrt{3}σe}{2mε_0}## which gives incorrect answer .

Is this the correct way to approach the problem ?

I would be grateful if somebody could help me with the problem.

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Tanya Sharma said:

## Homework Statement

An infinite dielectric sheet having charge density σ has a hole of radius R in it. An electron is released on the axis of the hole at a distance R√3 from the centre. What will be the velocity which it crosses the plane of sheet? (e = charge on electron and m = mass of electron)

## The Attempt at a Solution

Force at a distance x on the axis of the hole can be calculated by finding electric field at a distance x .Electric field is found by superimposing the electric field due to the sheet and the circular disk containing charge density -σ .

## E = \frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##mv\frac{dv}{dx} = -\frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##vdv = -\frac{σ}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

Now integrating under proper limits I get ## v^2 = \frac{\sqrt{3}σe}{2mε_0}## which gives incorrect answer .

Is this the correct way to approach the problem ?

I would be grateful if somebody could help me with the problem.

You certainly miss an R from the result. And you should show your work in detail...

ehild

Electric field due to infinite sheet having charge density σ = σ/2ε0

Electric field due to disk of radius R having charge density -σ at a distance x from the center of the disk on the axis = ##\frac{σ}{2ε_0}(1-\frac{x}{\sqrt{R^2+x^2}})##

Net electric field at a distance x , ## E = \frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##mv\frac{dv}{dx} = -\frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##vdv = -\frac{σ}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

##\int_{0}^{v} vdv = -\int_{\sqrt{3}R}^{0} \frac{σ}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

## v^2 = \frac{\sqrt{3}σe}{2mε_0}##

How did you cancel R from the integral?

ehild

1 person
ehild said:
How did you cancel R from the integral?

ehild

I apologize for the pathetic calculations .It's quite embarrassing for me

Thanks for the input.

Some "e"-s are also missing from your formulae. It would be nice to show your solution in a correct way. Think of the people who want to learn from the thread. What have you got at the end?

ehild

Electric field due to infinite sheet having charge density σ = σ/2ε0

Electric field due to disk of radius R having charge density -σ at a distance x from the center of the disk on the axis = ##\frac{σ}{2ε_0}(1-\frac{x}{\sqrt{R^2+x^2}})##

Net electric field at a distance x , ## E = \frac{σ}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##mv\frac{dv}{dx} = -\frac{σe}{2ε_0}\frac{x}{\sqrt{R^2+x^2}}##

##vdv = -\frac{σe}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

##\int_{0}^{v} vdv = -\int_{\sqrt{3}R}^{0} \frac{σe}{2mε_0}\frac{x}{\sqrt{R^2+x^2}}dx##

## [\frac{v^2}{2}]_0^v= -\frac{σe}{2mε_0}[\sqrt{R^2+x^2}]_{\sqrt{3}R}^0##

## v^2= \frac{σe}{mε_0}[2R-R]##

## v^2 = \frac{σeR}{mε_0}##

## v = \sqrt{\frac{σeR}{mε_0}}##

ehild...Does it look alright ?

If I were to approach this problem using energy conservation then too ,I would have to perform same integration ,as the right hand side is nothing but the change in the electric potential energy of the system .

Splendid!

ehild

## 1. What is an infinite sheet in electrostatics?

An infinite sheet in electrostatics is a theoretical construct used to simplify calculations and models in situations where there is a large, uniform distribution of electric charge. It is a perfectly flat surface with an infinite area that has a constant charge density throughout.

## 2. How is the electric field calculated for an infinite sheet?

The electric field for an infinite sheet is calculated using Gauss's Law, which states that the electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space. In this case, the electric field is constant and perpendicular to the sheet, and its magnitude is equal to the charge density divided by the permittivity of free space.

## 3. Can an infinite sheet have a non-zero net charge?

No, an infinite sheet cannot have a non-zero net charge. This is because if it did, the electric field would diverge to infinity, which violates the fundamental principles of electrostatics. An infinite sheet must have an equal amount of positive and negative charge to maintain a constant and finite electric field.

## 4. How does an infinite sheet affect the electric field of a nearby point charge?

An infinite sheet has a strong influence on the electric field of a nearby point charge. The electric field of the infinite sheet is constant and perpendicular to the sheet, so the presence of the sheet can change the direction and magnitude of the electric field from the point charge. This is due to the superposition principle, which states that the total electric field at a point is the sum of the individual electric fields from all nearby charges.

## 5. What are some real-world applications of infinite sheets in electrostatics?

Infinite sheets are used in various applications, such as in the design of parallel plate capacitors, which use two infinite sheets to store electric charge. They are also used in modeling the electric field of large, flat surfaces, such as conductive plates or mirrors. Additionally, infinite sheets are used in theoretical calculations to simplify complex electrostatic problems and to better understand the behavior of electric fields in different scenarios.

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