Is Direction Considered in Determining Resultant Electric Field Strength?

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v_pino
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How do I know that the resultant filed strength acts toward Y?

Thank you
 

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Definition of electric field strength:

"Field strength at a point O is the force experienced (Here force means both magnitude AND direction) by a charge of +1 Coulomb kept at point O due to all other charges."

Now, suppose a +1 C charge is kept at point O.

Due to -Q charge kept at point Z, the +1C charge at O will experience an attraction (opposite charges attract)

so due to Z, the direction will be towards Z. ( [tex]\longleftarrow[/tex])

similarly due to charge at X: ([tex]\longrightarrow[/tex])
These two are equal in magnitude but opposite in direction. So both will cancel each other.




The charge at W is positive, so it will repel the +1C charge kept at O. So force on +1C due to W will be in the upward direction.

Again, charge at Y is negative. So it will attract the +1C towards itself (up). So force due to Y is up.

Field due to charges at W and Y will add up to give a resultant field in the upward direction.
 
But this is only the case when we assume that it is +1C at O. If we assume that it is negative that the resultant field strength is downwards. How come we can be sure to make the assumtion that it is +1C?
 
v_pino said:
But this is only the case when we assume that it is +1C at O. If we assume that it is negative that the resultant field strength is downwards. How come we can be sure to make the assumtion that it is +1C?

Electric Field at a point is defined assuming that a +1 C charge is kept at that point. It is written in the standard scientific literature or something. We cannot take electric field to be the force experienced by -1 C charge.

Field strength at a point O is the force experienced by a charge of +1 Coulomb kept at point O due to all other charges.
This is a definition, no one can change it.


But yes, if they specify in the question that -1 C charge is kept at O, then it will not tell you to calculate field, they will tell you to calculate force on -1C instead. In that case, the resultant force will be downwards.
 
the biggest flaw in the way you are solving this question, as google spider said is that you are not at all accounting for the direction of the field. Work in terms of unit vectors [itex]\hat{i},~\hat{j}[/itex]. It'll be much easier.