Calling all Physics smarties - 2 Questions - 1 Best Answer - Help is APPRECIATED :D?

Question

I have 2 Questions, feel free to answer if you'd like. All help is appreciated. Unfortunately, I can only choose one best answer.

QUESTION ONE:
Two point charges, q1 = +20.0nC and q2 = +10.0 nC, are located on the x-axis at x = 0 and x = 1.00 m, respectively. Where on the x-axis is the electric field equal to zero?

QUESTION TWO:

Two point charges, q1 = +20.0nC and q2 = +10.0 nC, are located on the x-axis at x = 0 and x = 1.00 m, respectively. What is the magnitude of the electric field at the point x = 0.50 m, y = 0.50 m?

Once agian, ANY Help is appreciated! Thanks to those who take the time to help! :)

Answer 1

Question one:

Let a point between the charges be at x0. The distance from the 20nC charge is x0, the distance from the 10nC charge is 1-x0. The field from the first charge is K*20/x0^2. The field from the second charge is K*10/(1-x0)^2 in the opposite direction. The force is zero when the fields are equal and cancel out. Therefore

K*20/x0^2 = K*10/(1-x0)^2

2/x0^2 = 1/(1-x0)^2

2*(1-x0^2) = x0^2

2 - 4*x0 + 2*x0^2 = x0^2

x0^2 - 4*x0 + 2 = 0

Solve using quadratic equation; the only solution for x0<1 is x0 = 0.586

Question two:

The distance from each charge to the given point is √[.5^2+.5^2] = √[.25+.25] = √.5

The field from the charges is then q1/(4πe0*0.5) and q2/(4πe0*0.5). You add the x and y components of each field, noting that the angle of the field from the x-axis is 45º

x-components: q1/(2πe0)*cos(45º) - q2/(2πe0)*cos(45º)

y-components: [q1/(2πe0)]*sin(45º) + [q2/(2πe0)]*sin(45º)

Since sin(45º) = cos(45º) = √2/2

x-comp of E is √2/(4πe0)*(q1 - q2)
y comp of E is √2/(4πe0)*(q1 + q2)

The sqrt of sum of squares of x and y components is the magnitude of E, and the angle is arctan (y-comp/x-comp)

Answer 2

1: x=10/[(1/2)^(1/2)+1]
2: 5^(1/2)180 and its degree is 45 + arcsin ([5^(1/2)]/5)with x-axis