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Chapter 11: Problem 1541

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

Expert verified
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

01

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 • \(QR = 1 \times 10^{-2} m\).
02

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 \(E_q\): \[E_q = \frac{9\times10^9 \cdot 2\times10^{-6}}{(1\times10^{-2})^2} = 1.8\times10^4 NC^{-1}\]
03

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.
04

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: \[E_r = \sqrt{E_{rx}^2 + E_{ry}^2} = \sqrt{(9\times10^3)^2 + (1.8\times10^4)^2} = 2.05\times10^4 NC^{-1}\]
05

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).

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Most popular questions from this chapter

An inclined plane making an angle of \(30^{\circ}\) with the horizontal is placed in an uniform electric field \(E=100 \mathrm{Vm}^{-1}\). A particle of mass \(1 \mathrm{~kg}\) and charge \(0.01 \mathrm{c}\) is allowed to slide down from rest from a height of \(1 \mathrm{~m} .\) If the coefficient of friction is \(0.2\) the time taken by the particle to reach the bottom is $\ldots \ldots . .$ sec (A) \(2.337\) (B) \(4.337\) (C) 5 (D) \(1.337\)

The electric Potential \(\mathrm{V}\) at any Point $0(\mathrm{x}, \mathrm{y}, \mathrm{z}\( all in meters \))\( in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}\( volt. The electric field at the point \)(1 \mathrm{~m}, 0.2 \mathrm{~m})\( in volt meter is \)\ldots \ldots .$ (A) 8 , along negative \(\mathrm{x}\) - axis (B) 8 , along positives \(\mathrm{x}\) - axis (C) 16 , along negative \(\mathrm{x}\) -axis (D) 16 , along positives \(\mathrm{x}\) -axis

Three concentric spherical shells have radii a, \(b\) and \(c(a

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the $\ldots \ldots\( charged particles in free space is (given \)\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)$ (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

A point charge \(q\) is situated at a distance \(r\) from one end of a thin conducting rod of length \(\mathrm{L}\) having a charge \(\mathrm{Q}\) (uniformly distributed along its length). The magnitude of electric force between the two, is ...... \((\mathrm{A})[(2 \mathrm{kqQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (B) \([(\mathrm{kq} \mathrm{Q}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (C) \([(\mathrm{kqQ}) / \mathrm{r}(\mathrm{r}-\mathrm{L})]\) (D) \([(\mathrm{kQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\)
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