# Electrostatics problem involving infinite sheet

1. Mar 10, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

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)

2. Relevant equations

3. 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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Last edited: Mar 10, 2014
2. Mar 10, 2014

### ehild

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

ehild

3. Mar 10, 2014

### Tanya Sharma

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

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

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

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

$v^2 = \frac{\sqrt{3}σe}{2mε_0}$

4. Mar 10, 2014

### ehild

How did you cancel R from the integral?

ehild

5. Mar 10, 2014

### Tanya Sharma

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

Thanks for the input.

6. Mar 11, 2014

### ehild

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

7. Mar 11, 2014

### Tanya Sharma

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 .