Electric field insulating rod Question

[ToonBlue]

Homework Statement

Homework Equations

The Attempt at a Solution

This is problem I have found online and this is the part I don't understand how they come up with this.

How did they get this for the x component since y component cancel out due to being symmetry ?

 

[Antonius]

I think they are trying to find dE not E...

 

[ToonBlue]

The above is the formula given in my lecture note .

In the solution provided in the problem , it is given as

which means that

is the unit direction of the component.

In this case , since there is no y component (symmetry) , it has only x component but how do they get

?

 

[Antonius]

The y component will cancel out I think when you take the Integral over all points towards point O. I believe the formula ( at the end ) provided in the solution is trying to calculate the dE at point O.
Wait for more qualified members to reply. I am taking Electricity and Magnetism course right now :)

 

[NascentOxygen]

The diagram in your first post shows you dE; it's the field you'll see at O due to the charge contained in the short length, dl. The electric field at any point has both a magnitude (illustrated by the arrow having a length) and a direction (shown by the angle of arrow dE to the reference axes).

It's not until you sum the fields due to multiple fragments of charge will you see cancellations, this entails integrating dE for all of the line charge.

 

[rude man]

Homework Statement

View attachment 88284

View attachment 88288

Homework Equations

The Attempt at a Solution

This is problem I have found online and this is the part I don't understand how they come up with this.
View attachment 88285
How did they get this for the x component since y component cancel out due to being symmetry ?
View attachment 88286

The y component cancels out only after you have integrated dE.

 

[ToonBlue]

May I know why does the dE point away and not toward the dL?

 

[rude man]

May I know why does the dE point away and not toward the dL?[/QUOThe arc L is negatively charged. The direction of the E field is the direction of the force on a positive unit of charge. So the E field is directed towards the arc L, as you say.
If you got the impression it's directed away from L it's probably because of the picture. However, the picture is still right since the answer is negative.