Question

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

Short answer

The magnitude and direction of the resultant electric field at R are \(2.05\times10^4 NC^{-1}\) and \(34.7^{\circ}\), respectively.

Step-by-step solution

Calculate individual electric fields

First, we will calculate the magnitude of the electric fields at point R due to the charges at points P and Q using Coulomb's law. The electric field at point R due to a point charge at P is: \[E_p = \frac{k \cdot Q_p}{(PR)^2}\] and the electric field at point R due to a point charge at Q is \[E_q = \frac{k \cdot Q_q}{(QR)^2}\] where • \(k = 9\times10^9 Nm^2C^{-2}\) (Coulomb's constant), • \(Q_p = 4 \times 10^{-6} C\), • \(Q_q = 2 \times 10^{-6} C\), • \(PR = 2 \times 10^{-2} m\), and • \(QR = 1 \times 10^{-2} m\).

Calculate magnitudes of electric fields

Now, we can calculate the magnitudes of electric fields: For \(E_p\): \[E_p = \frac{9\times10^9 \cdot 4\times10^{-6}}{(2\times10^{-2})^2} = 9\times10^3 NC^{-1}\] For \(E_q\): \[E_q = \frac{9\times10^9 \cdot 2\times10^{-6}}{(1\times10^{-2})^2} = 1.8\times10^4 NC^{-1}\]

Calculate the components of the electric fields

Now, we need to calculate the components of the electric fields at point R. Let's assume the positive x-axis is along PR, and the positive y-axis is along QR. From the given data, we can see that: • \(E_{px} = E_p\), since it is along the x-axis. • \(E_{py} = 0\), since there is no electric field component along the y-axis. • \(E_{qx} = 0\), since there is no electric field component along the x-axis. • \(E_{qy} = E_q\), since it is along the y-axis.

Calculate the resultant electric field

Now, we can calculate the resultant electric field by adding the components of the electric fields: • \(E_{rx} = E_{px} + E_{qx} = E_p + 0 = 9\times10^3 NC^{-1}\) • \(E_{ry} = E_{py} + E_{qy} = 0 + E_q = 1.8\times10^4 NC^{-1}\) The magnitude of the resultant electric field is: \[E_r = \sqrt{E_{rx}^2 + E_{ry}^2} = \sqrt{(9\times10^3)^2 + (1.8\times10^4)^2} = 2.05\times10^4 NC^{-1}\]

Calculate the direction of the resultant electric field

Now we can calculate the angle θ between the x-axis and the electric field: \[tan \theta = \frac{E_{ry}}{E_{rx}} = \frac{1.8\times10^4}{9\times10^3}\] Therefore, \[\theta = arctan\left(\frac{1.8\times10^4}{9\times10^3}\right) \approx 34.7^{\circ}\] So the magnitude and direction of the resultant electric field at R is \(E_r \approx 2.05\times10^4 NC^{-1}\) and \(\theta \approx 34.7^{\circ}\), which matches option (C).