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).

Key concepts

Coulomb's Law

Coulomb's Law is the fundamental principle that describes the interaction between two point charges. According to Coulomb's Law, the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this is represented as:\[ F = \frac{k \cdot Q_1 \cdot Q_2}{r^2} \]Where:

• \( F \) is the magnitude of the force between the charges,
• \( k \, (9 \times 10^9 \, \text{N m}^2/ \text{C}^2) \) is Coulomb's constant,
• \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
• \( r \) is the distance between the charges.

The force is attractive if the charges are opposite, and repulsive if the charges are the same. This law is crucial when calculating electric fields because it helps determine the magnitude of fields generated by point charges.

Point Charges

In physics, a point charge refers to an idealized charge that is considered to occupy no space and therefore acts from a single point in space. This simplification helps in solving problems related to electric fields and forces, as actual physical measurements may involve complex charge distributions.

For example, in the given exercise, the charges at points P and Q are assumed to be point charges of magnitudes \( 4 \, \mu \mathrm{C} \) and \( 2 \, \mu \mathrm{C} \) respectively. This simplification allows us to apply Coulomb's Law effectively to calculate the electric fields they produce. By treating these as point charges, we are essentially saying that all of the charge is concentrated at a single point at their respective locations within the setup.

Electric Field Components

The electric field is a vector quantity characterized by both direction and magnitude. At any given point in space, the resultant electric field due to multiple charges can be computed by dividing the field into components. This allows easier manipulation and combination of vectors, particularly in scenarios involving vector addition.
The component approach:

• Identify the axis directions usually given in the problem (e.g., x and y-axes).
• Resolve each electric field into its components along these axes.

In the step by step solution, the field due to the charge at point P was purely along the x-axis making its y-component zero. Conversely, the field due to the charge at point Q was purely along the y-axis. By using this strategy, fields are easier to combine to find the resultant field.

Resultant Electric Field

The resultant electric field at a given point is the vector sum of all individual electric field components influencing that point. Once the components of each electric field are identified, the resultant field is derived using vector addition.
To find the resultant:

• Add all x-components of the fields to get \( E_{rx} \).
• Add all y-components of the fields to get \( E_{ry} \).
• The magnitude \( E_r \) of the resultant field is given by: \[ E_r = \sqrt{E_{rx}^2 + E_{ry}^2} \]
• To find the direction, calculate the angle \( \theta \) with respect to the x-axis using the \[ \tan \theta = \frac{E_{ry}}{E_{rx}} \]

This calculation yields both the magnitude and direction of the electric field, providing a complete description of the field at that location. In our exercise, the calculated resultant field magnitude and direction matched with option (C).