How Does Charge Distribution Affect Electric Field Magnitude?

[umagongdi]

Homework Statement

The x-y plane contains point charge with magnitudes and positions as follows:

A charge of Q lies at co-ordinates (0, a)
A charge of -Q lies at co-ordinates (0,-a)

Show that the magnitude of the electric field at (b, 0) is given by

E = Qa / [2πε0(a²+b²)^(3/2)]

ε0 = Epsilon nought = 8.85 x10^-12

Homework Equations

E = Q/ 4πε0r²

The Attempt at a Solution

E due to Q is

E1 = Q/ 4πε0r²
E1 = Q/ 4πε0(a²+b²)

Therefore the vector E'

E' = E1*Cos(x) i^ + E1*Sin(x) j^
E' = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^

E due to -Q is

E2 = Q/ 4πε0r²
E2 = Q/ 4πε0(a²+b²)

Therefore the vector E''

E'' = E2*Cos(x) i^ + E2*Sin(x) j^
E'' = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^

Vector E = E' + E''
E = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^
+ QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^
E = QCos(x) / 2πε0(a²+b²) i^ + QSin(x)/ 2πε0(a²+b²) j^

Magnitude of E
E = sqrt{Q²Cos²(x) / [2πε0(a²+b²)]² + Q²Sin²(x)/ [2πε0(a²+b²)]²}

E = Q / 2πε0(a²+b²)

As you can see its close lol, need Q*a not Q by itself and (a²+b²) should be(a²+b²)^(3/2)?

 

[Matterwave]

remember that one charge is +Q and another is -Q. Also remember that electric fields are vectors and therefore direction matters, so your use of "cos(x)" and "sin(x)" in both equations begs the question "what is x referring to here?" So, my advice is to draw a diagram. It will make the question much simpler.

 

[umagongdi]

Sorry i forgot to explain what x is. After drawing a diagram with resolved forces at (b,0), x would be the angle between the resultant E fields and the x axis. I am not sure which signs need to be negative any clues/advice/lecture will be appreciated very much.