# What is the net electric field?

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Given that #Q_1=7xx10^(-6)C# is located at the origin and #Q_2=5xx10^(-6)C# is located #0.3"m"# to the right of #Q_1# , what is the net electric field at a point #P# located #0.4"m"# above #Q_1# ?

Given that

##### 1 Answer

#### Answer:

#### Explanation:

The **electric field** of a point charge is given by:

#vecE=kabs(q)/r^2# where

#k# is the electrostatic constant,#q# is the magnitude of the charge, and#r# is the radius from the charge to the specified point

The *net* electric field at point **vector sum** of electric fields

#(E_x)_(n et)=sumE_x=E_(x1)+E_(x2)#

#(E_y)_(n et)=sumE_y=E_(y1)+E_(y2)#

#E_(n e t)=sqrt((E_x)^2+(E_y)^2)#

So, in order to find the net electric field at point *positive* charges create electric fields with vectors that point *away* from them.

I will only draw in a couple of the vectors—those that are relevant to the problem—but as in the above picture, the field lines point out (or in) in every direction from the charge.

**Diagram:**

The electric field vector originating from **only a perpendicular component**, so we will not have to worry about breaking this one up.

Therefore,

#E_1=kabs(Q_1)/r^2#

#=((8.99*10^9("N"*"m")/"C"^2)(7*10^-6"C"))/(0.4"m")^2#

#=393312.5" N"//"C"#

In order to calculate

#x^2+y^2=r^2#

#=>r=sqrt(x^2+y^2)#

#=sqrt(0.3^2+0.4^2)#

#=0.5#

We can now calculate

#E_2=kabs(Q_2)/r^2#

#=((8.99*10^9("N"*"m")/"C"^2)(5*10^-6"C"))/(0.5"m")^2#

#=17980000" N"//"C"#

This field vector occurs at an angle relative to

We have:

#(E_2)_x=E_2cos(theta)#

#(E_2)y=E_2sin(theta)#

Before we calculate the components, we'll have to find the angle. We can do this using the arctangent function, since we have both of the triangle's side lengths.

#tan(theta)=y/x#

#=>theta=arctan(y/x)#

#=arctan(0.4/0.3)#

#=53.13^o#

Therefore:

#(E_2)_x=(17980000" N"//"C")*cos(53.13^o)#

#=10788025.7 " N"//"C"#

#(E_2)_y=(17980000" N"//"C")*sin(53.13^o)#

#=14383980.73" N"//"C"#

Note that because all of these components occur above the positive x-axis and to the right of the origin, all of them have positive values. This may not always be the case, so be sure to keep track of your signs.

We now have:

#E_x=10788025.7 " N"//"C"#

#E_y=393312.5" N"//"C" + 14383980.73" N"//"C"#

#=14777293.23" N"//"C"#

We can now find the net electric field at

#E_(n e t)=sqrt((E_x)^2+(E_y)^2)#

#=sqrt((10788025.7)^2+(14777293.23)^2)#

#=18296171.56" N"//"C"#

This is a fairly large quantity, so we would likely express it in scientific notation as