Force between charged particle and conducting cylinder

[Bling Fizikst]

Homework Statement

refer to image

Relevant Equations

refer to image

Since $x$ is larger than the linear dimensions of the cylinder, hence it can be approximated to be a point dipole . $$\vec{F}=\left(\vec{\nabla}\cdot \vec{p}\right)\vec{E}$$ In our case : $$\vec{F}=p_x\frac{\partial E_x}{\partial x}$$ $$F=\frac{-2kQp_x}{x^3}$$ Of course , we can assume some charges $-q , +q$ being induced on the sides of the cylinder in order to find $p_x$ but i don't know how .

 

[haruspex]

Homework Statement: refer to image
Relevant Equations: refer to image

View attachment 346310
Since $x$ is larger than the linear dimensions of the cylinder, hence it can be approximated to be a point dipole . $$\vec{F}=\left(\vec{\nabla}\cdot \vec{p}\right)\vec{E}$$ In our case : $$\vec{F}=p_x\frac{\partial E_x}{\partial x}$$ $$F=\frac{-2kQp_x}{x^3}$$ Of course , we can assume some charges $-q , +q$ being induced on the sides of the cylinder in order to find $p_x$ but i don't know how .
F=\px∂Ex∂x=2kQpx3

If you are looking at the conducting case, what is the net field in the conductor? What does that tell you about the induced charges?

 

[Bling Fizikst]

If you are looking at the conducting case, what is the net field in the conductor? What does that tell you about the induced charges?

I did try with gauss law but got something that i can't really make sense of maybe .

So , i considered the entire cylinder to be a gaussian surface . Let the electric field entering the face be $$ E_1=\frac{kQ}{x^2}$$ and the field leaving the other side of the face be $$E_2=\frac{kQ}{(x+d)^2}$$ where $d$ is the length of the cylinder .

Applying gauss law : $$-E_1 A+E_2 A =\frac{q_{\text{induced}}}{\epsilon_{\circ}}$$ Substituting and applying approximations considering $x>>d$ , i got $$ q_{\text{induced}} = -\frac{kQ(2V)}{x^3}$$ But shouldn't the net charged enclosed by the cylinder be $-q+q=0??$

 

[haruspex]

I did try with gauss law but got something that i can't really make sense of maybe .

So , i considered the entire cylinder to be a gaussian surface . Let the electric field entering the face be $$ E_1=\frac{kQ}{x^2}$$ and the field leaving the other side of the face be $$E_2=\frac{kQ}{(x+d)^2}$$ where $d$ is the length of the cylinder .

Applying gauss law : $$-E_1 A+E_2 A =\frac{q_{\text{induced}}}{\epsilon_{\circ}}$$ Substituting and applying approximations considering $x>>d$ , i got $$ q_{\text{induced}} = -\frac{kQ(2V)}{x^3}$$ But shouldn't the net charged enclosed by the cylinder be $-q+q=0??$

I think the apparent contradiction is explained by the small part of the field that emanates from the sides of the cylinder.
But you still have not said whether this is the conductor case or the insulator case.
If conductor, what must the net field in the cylinder be?

 

[Bling Fizikst]

I think the apparent contradiction is explained by the small part of the field that emanates from the sides of the cylinder.
But you still have not said whether this is the conductor case or the insulator case.
If conductor, what must the net field in the cylinder be?

Yeah i was dealing with the conductor case . The net electric field inside a conductor is zero .

 

[haruspex]

Yeah i was dealing with the conductor case . The net electric field inside a conductor is zero .

So what field must the induced charge produce in the cylinder?

 

[Bling Fizikst]

So what field must the induced charge produce in the cylinder?

The induced electric field will cancel the net external field entering the cylinder . So , i think the net external field should be $$E_2-E_1=-\frac{kQ(2d)}{x^3}=-E_{\text{ind}}=-\frac{kq}{d^2}?$$

 

[haruspex]

The induced electric field will cancel the net external field entering the cylinder . So , i think the net external field should be $$E_2-E_1=-\frac{kQ(2d)}{x^3}=-E_{\text{ind}}=-\frac{kq}{d^2}?$$

I agree with the words but cannot relate them to your equations. Which of those terms is the external field? (Why would that term involve d?)

 

[Bling Fizikst]

I agree with the words cannot relate them to your equations. Which of those terms is the external field? (Why would that term involve d?)

i am kind of stuck at this point . i believe $E_{\text{ext}}=\frac{kQ}{x^2}?$

I tried to apply gauss law a bit differently . I took a gaussian pill box that is 'half' the length of the original cylinder . Which should give me the induced charge $$q=\frac{kQV\epsilon_{\circ}}{x^3}$$ Plugging this in the dipole moment $$p=qd$$ I got : $$F=\frac{Q^2 V d}{8\pi^2\epsilon_{\circ} x^6}$$ But don't know how to eliminate $d$

 

[haruspex]

i am kind of stuck at this point . i believe $E_{\text{ext}}=\frac{kQ}{x^2}?$

Right.

I tried to apply gauss law a bit differently . I took a gaussian pill box that is 'half' the length of the original cylinder . Which should give me the induced charge $$q=\frac{kQV\epsilon_{\circ}}{x^3}$$

I'm not seeing how you got that.

Suppose the cylinder has cross section A=V/d.
If the induced charges in the cylinder are q and -q, what field do those produce within the cylinder? What equation does that allow you to write?

 

[Bling Fizikst]

Right.

I'm not seeing how you got that.

Suppose the cylinder has cross section A=V/d.
If the induced charges in the cylinder are q and -q, what field do those produce within the cylinder? What equation does that allow you to write?

A cylinder's cross section? makes me think about capacitors! Electric field between two capacitor plates would be , $$E=\frac{q}{A\epsilon_{\circ}}$$ Since this electric field will cancel the external electric field so : $$\frac{q}{A\epsilon_{\circ}} = \frac{kQ}{x^2}$$ $$\implies q=\frac{kQA\epsilon_{\circ}}{x^2} \implies p=qd=\frac{kQV\epsilon_{\circ}}{x^2}$$ Plugging this value of $q$ in the force equation : $$F=-\frac{Q^2 V}{8\pi^2\epsilon_{\circ} x^5}$$