Question

\(\mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and a \(-24.00-\mathrm{nC}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?

Short answer

Short Answer: The direction of the electric field at the origin is approximately -12.5°, meaning it points downwards from the positive x-axis towards the negative y-axis.

Step-by-step solution

Find the electric field due to the \(+48.00\) nC charge

The electric field \(E\) exerted by a point charge \(q\) has a magnitude of: \(E = \frac{k \cdot |q|}{r^2}\) where \(r\) is the distance from the charge and \(k = 8.9875 \times 10^9 \mathrm{\ N\cdot m^2 / C^2}\) is Coulomb's constant. Since the charge is at the point \((4, 0)\), we can find its distance from the origin using the Pythagorean theorem: \(r_1 = \sqrt{(4 - 0)^2 + (0 - 0)^2} = 4 \mathrm{\ m}\). Now we can find the magnitude of the electric field due to the \(+48.00\) nC charge: \(E_1 = \frac{8.9875\times 10^9 \mathrm{\ N\cdot m^2 / C^2} \cdot 48.00\times 10^{-9} \mathrm{\ C}}{(4 \mathrm{\ m})^2} = 27.00 \mathrm{\ kN/C}\).

Find the electric field due to the \(-24.00\) nC charge

Similarly, to find the magnitude of the electric field due to the \(-24.00\) nC charge, we first calculate its distance from the origin: \(r_2 = \sqrt{(0 - 0)^2 + (-6 - 0)^2} = 6 \mathrm{\ m}\). Now we can find the magnitude of the electric field due to the \(-24.00\) nC charge: \(E_2 = \frac{8.9875\times 10^9 \mathrm{\ N\cdot m^2 / C^2} \cdot |-24.00\times 10^{-9} \mathrm{\ C}|}{(6 \mathrm{\ m})^2} = 6.00 \mathrm{\ kN/C}\).

Find the net electric field at the origin

Since both electric fields are acting in the same plane, we can compute the x-component and y-component of each electric field as follows: \(E_{1x} = E_1\cos\theta_1 = 27.00 \mathrm{\ kN/C}\) (as it is on the x-axis, the angle is 0°) \(E_{1y} = E_1\sin\theta_1 = 0\) (no y-component) \(E_{2x} = E_2\cos\theta_2 = 0\) (no x-component) \(E_{2y} = E_2\sin\theta_2 = -6.00 \mathrm{\ kN/C}\) (as it is on the y-axis) Now we sum up the x- and y-components to get the net electric field at the origin: \(E_{netx} = E_{1x} + E_{2x} = 27.00 \mathrm{\ kN/C} + 0 = 27.00 \mathrm{\ kN/C}\) \(E_{nety} = E_{1y} + E_{2y} = 0 + (-6.00 \mathrm{\ kN/C}) = -6.00 \mathrm{\ kN/C}\) Finally, to find the direction of the net electric field, we calculate the angle \(\alpha\): \(\alpha = \arctan{\frac{E_{nety}}{E_{netx}}} = \arctan{\frac{-6.00 \mathrm{\ kN/C}}{27.00 \mathrm{\ kN/C}}} \approx -12.5°\) The direction of the electric field at the origin is approximately -12.5°, which indicates that the electric field is pointing downwards from the positive x-axis towards the negative y-axis.