Find strength and direction of electric field

[sheepcountme]

Homework Statement

There is a charge of -4.05mC at (3,0)m and a charge of +3.85mC at (10,0)m. And I have to find the strength and direction of the field at (2.5,7)m. (Direction to be in degrees).

Homework Equations

E=(kq)/r^2

The Attempt at a Solution

I've found the fields to be 7.40x10^5 N/C for the first and 3.28x10^5 N/C for the second on the point.

And I'm stuck from here. I've started figuring out all sorts of angles but I just get swamped. Could someone please nudge me in the right direction?

 

[rl.bhat]

Can you show your calculations of the fields?

 

[sheepcountme]

Sure thing, for the first:
(9x10⁹)(4.05x10^-3 / 49.25

The 49.25 came from the distance (2.5-3)²+(7-0)² (the sqrt of this and then squaring again would cancel each other so I didn't bother with it)

So, that ended up being (3.645x10⁷)/49.25 equaling 740,102=7.40x10⁵

For the second:
(9x10⁹)(3.85x10^-3)/105.25

The 105.25 coming from (2.5-10)²+(7-0)²) again ignoring the sqrt squared and getting 56.25+49=105.25

And so, (3.465x10⁷)/105.25 = 328,361 = 3.28x10⁵

 

[rl.bhat]

OK. Now the electric field E1 due to the first charge is towards the charge and the field E2 due to the second charge is away from the charge. To find the resultant we must know the angels made by these fields with horizontal.
For the Ε1, tanθ1 = 7/0.5 and for E2, tanθ2 = 7/7.5. Find θ1 and θ2.

Now take the horizontal and vertical components of E1 and E2 and find the resultant E.

 

[rwisz]

I would approach this problem by first drawing a diagram to visualize the situation. From the given information, I can see that there are two charges, one negative and one positive, located at specific coordinates. The point of interest is located at (2.5,7)m, which is not directly in between the two charges. This means that the electric field at this point will be affected by both charges.

Next, I would use the equation E=(kq)/r^2 to calculate the electric field strength at the point of interest due to each charge. This will give me two separate values, one for the electric field strength due to the negative charge and one for the electric field strength due to the positive charge.

To find the total electric field strength at the point of interest, I would use the principle of superposition. This means that I would add the two electric field strengths together, taking into account their directions. The direction of an electric field is given by the direction of the force it would exert on a positive test charge placed at that point. Since a positive test charge would be repelled by a negative charge and attracted by a positive charge, I would assign the direction of the electric field due to the negative charge as being away from the charge and the direction of the electric field due to the positive charge as being towards the charge.

Finally, to find the direction of the total electric field, I would use trigonometry to find the angle between the total electric field vector and the x-axis. This angle would be the direction of the electric field in degrees.

In summary, to find the strength and direction of the electric field at (2.5,7)m, you need to calculate the electric field strength at this point due to each individual charge, add them together, and then use trigonometry to find the direction in degrees.