Is force, which we show, right on test charge

[rahaverhma]

As we draw the electric field lines due to 2 equal magnitude and same nature charges. But when we solve the question concerning the same case we take the resultant of forces due to both charges at a test charge in such a manner that the 2nd charge(let I put the test charge on the left side i.e. the region of 1st charge) is also exerting the force but it can't because the electric field line due to that charge actually never crossed the margin bisecting the line joining the 2 charges. Is it right?

 

[berkeman]

As we draw the electric field lines due to 2 equal magnitude and same nature charges. But when we solve the question concerning the same case we take the resultant of forces due to both charges at a test charge in such a manner that the 2nd charge(let I put the test charge on the left side i.e. the region of 1st charge) is also exerting the force but it can't because the electric field line due to that charge actually never crossed the margin bisecting the line joining the 2 charges. Is it right?

Could you please post a diagram of the situation you are asking about? Is it something like this?

https://image.slidesharecdn.com/ele...1/95/electric-fields-11-728.jpg?cb=1237897460

 

[rahaverhma]

Could you please post a diagram of the situation you are asking about? Is it something like this?

https://image.slidesharecdn.com/ele...1/95/electric-fields-11-728.jpg?cb=1237897460
View attachment 222668

Yes, sir. But I'm not putting test charge on the line because I know there I will be wrong .

 

[berkeman]

What line? What does your diagram look like? You can use the UPLOAD button in the lower right of the Reply box to Upload a PDF or JPEG of your diagram...

 

[rahaverhma]

Yes, sir. But I'm not putting test charge on the line because I know there I will be wrong .

Sorry, I don't know

 

[rahaverhma]

What line? What does your diagram look like? You can use the UPLOAD button in the lower right of the Reply box to Upload a PDF or JPEG of your diagram...

https://www.google.co.in/search?q=E...gB&biw=360&bih=560&dpr=2#imgrc=RLgZKy3iG65q3M:
And
https://www.google.co.in/search?cli...j0i67j30i10.gSZuco4hX1w=#imgrc=CFk_H05Q_N2EyM:
And the magnitude and polarity of charges are same.

 

[Dale]

when we solve the question concerning the same case we take the resultant of forces due to both charges at a test charge

Yes, that is always the case. The field and therefore the force is linear, meaning that the force due to both charges is the sum of the forces due to each charge individually.