Electric field strength and direction problem (need help)

[ks81]

Homework Statement

Electric field strength and direction problem

Relevant Equations

E= k * q/r^2

E1= 9.0 x 10^9 * 10 x 10^-9 / (0.045)^2 = 45000. x cos(153.43)=
E2= 9.0 x 10^9 * 10 x 10^-9 / (0.02)^2= 22500 x cos(180)=
E3= 9.0 x 10^9 * 5.0 x 10^-9 / (0.04)^2= 28125 x cos(90)= ?

Enet= ?

 

[Delta2]

Please write a more detailed attempt at solution.

If I understand correctly at this attempt you try to calculate the three y-components of the three electric fields due to each of the three charges.
You must also write the three x-components and then sum the three y-components to calculate a total y-component, and similarly to calculate a total x-component.

 

[haruspex]

Further to @Delta2's reply, use commonsense to check your signs. It is very easy to get those wrong. E.g., which way should the field from the lower right charge point, up the page or down? (I'm not saying have that wrong, just that it is worth checking.)
Also, there is rarely any merit in finding the numeric value of an angle. You can find their trig functions with simple geometry.

 

[kuruman]

In addition to @haruspex's and @Delta2's replies, it seems that you are magnitudes of vectors with their components. For example

E1= 9.0 x 10^9 * 10 x 10^-9 / (0.045)^2 = 45000. x cos(153.43)=

The number .0 x 10^9 * 10 x 10^-9 / (0.045)^2 is equal to about 45000 and is equal to the magnitude of E1. Where did cos(153.43) come from? If you meant to write the x or y component of E1, then put down
E1x=E1*(some trig function)
E1y=E1*(some other trig function)
Otherwise you will probably get confused about what it is that you are calculating.

 

[Delta2]

Where did cos(153.43) come from?

I think $\theta=26.57$ is the angle of the diagonal with one of the sides of the parallelogram with sides 2 and 4. for some reason he takes $cos(180-\theta)$ not sure if this gives the y-component of the electric field of the charge at the bottom left corner. I thought so at the moment I was writing post #2.

 

[kuruman]

I think $\theta=26.57$ is the angle of the diagonal with one of the sides of the parallelogram with sides 2 and 4. for some reason he takes $cos(180-\theta)$ not sure if this gives the y-component of the electric field of the charge at the bottom left corner. I thought so at the moment I was writing post #2.

That may be so, however this

9.0 x 10^9 * 10 x 10^-9 / (0.045)^2 = 45000. x cos(153.43)

is not correct unless cos(153.43) = 1 which is clearly not the case. That is why I suggested that OP starts with E1x or E1y on the LHS rather than the magnitude E1.

 

[nasu]

Before even starting the calculations you should draw the forces or fields on the figure and label them. Don't expect people to guess what do you mean by E1, E2, E3 etc. As it is, they can be anything. Are these components, magnitudes? Without axes, how do we know which is x and which is y component?