Two electric charges, \(q_{1}=+20.0 \mathrm{nC}\) and \(q_{2}=+10.0 \mathrm{nC}\) are located on the \(x\) -axis at \(x=0 \mathrm{m}\) and \(x=1.00 \mathrm{m}\) respectively. What is the magnitude of the electric field at the point $x=0.50 \mathrm{m}, y=0.50 \mathrm{m} ?$

We're given the positions of charges \(q_{1}\) and \(q_{2}\) as \(x=0\) and \(x=1\) on the x-axis, respectively. The point we want to find the electric field at is \((0.50,0.50)\). The position vectors of the charges and the point can be written as: \(\vec{r}_{1} = (0, 0)\) \(\vec{r}_{2} = (1, 0)\) \(\vec{r}_{p} = (0.50, 0.50)\)

The distance vectors \(\vec{d}_{1}\) and \(\vec{d}_{2}\) from the charges to the point can be calculated by subtracting position vectors of charges from the position vector of the point: \(\vec{d}_{1} = \vec{r}_{p} - \vec{r}_{1} = (0.50, 0.50)\) \(\vec{d}_{2} = \vec{r}_{p} - \vec{r}_{2} = (-0.50, 0.50)\)

In order to calculate the magnitudes of the electric field vectors, we need to calculate the magnitudes of the distance vectors first: \(|\vec{d}_{1}| = \sqrt{(0.50)^2 + (0.50)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.50}\) \(|\vec{d}_{2}| = \sqrt{(-0.50)^2 + (0.50)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.50}\)

Using Coulomb's law, the electric field vectors due to each charge can be calculated as: \(\vec{E}_{1} = k \frac{q_{1}}{|\vec{d}_{1}|^2} \frac{\vec{d}_{1}}{|\vec{d}_{1}|}\) \(\vec{E}_{2} = k \frac{q_{2}}{|\vec{d}_{2}|^2} \frac{\vec{d}_{2}}{|\vec{d}_{2}|}\) Where \(k = 8.99 \times 10^9 N m^2/C^2\), the electrostatic constant. Substituting the values, we get: \(\vec{E}_{1} = 8.99 \times 10^9 \frac{20\times10^{-9}}{(0.50)^2} (\frac{0.50}{\sqrt{0.50}}, \frac{0.50}{\sqrt{0.50}})\) \(\vec{E}_{2} = 8.99 \times 10^9 \frac{10\times10^{-9}}{(0.50)^2} (-\frac{0.50}{\sqrt{0.50}}, \frac{0.50}{\sqrt{0.50}})\) Calculating the electric field vectors: \(\vec{E}_{1} = (25.45 \times 10^3, 25.45 \times 10^3) N/C\) \(\vec{E}_{2} = (-14.22 \times 10^3, 14.22 \times 10^3) N/C\)

Adding the electric field vectors due to \(q_{1}\) and \(q_{2}\), we get the net electric field vector: \(\vec{E}_{net} = \vec{E}_{1} + \vec{E}_{2} = (11.23 \times 10^3, 39.67 \times 10^3) N/C\)

Finally, finding the magnitude of the net electric field vector: \(|\vec{E}_{net}| = \sqrt{(11.23 \times 10^3)^2 + (39.67 \times 10^3)^2} = 41.86 \times 10^3 N/C\) The magnitude of the electric field at point \((0.50, 0.50)\) is \(41.86 \times 10^3 N/C\).