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Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{~cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{~cm} .\) (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\vec{E}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\). Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\), expressed in unit vector form.

Short Answer

Expert verified
The electric fields at point \(P\) are \(E1 = -28.0875 \hat{j} N/C\) and \(E2 = 19.4328 \hat{i} + 25.9104 \hat{j} N/C\), and the resultant field at \(P\) equals to \(E_{tot} = 19.4328 \hat{i} - 2.1771 \hat{j} N/C\).

Step by step solution

01

Define the Problem and Parameters

Charge \(q1\) is at the origin \(O(0,0)\) and charge \(q2\) is at \(A(3,0)\) on the \(x\)-axis. Point \(P\) is at \(B(0,4)\) on the \(y\)-axis. Charge \(q1\) is -5nC and \(q2\) is +3nC. Distances are in centimeters, but needs to be converted to meters.
02

Calculate the Distances

Calculate the distance \(r1\) from \(O\) to \(P\), and \(r2\) from \(A\) to \(P\). By using the Pythagorean theorem, the \(r1 = OP = 4cm = 0.04m\) and \(r2 = AP = \sqrt{(0.03)^2 + (0.04)^2} = 0.05m\).
03

Calculate the Electric Field due to Point Charges q1 and q2

Next, use Coulomb's Law to calculate electric fields \(E1\) and \(E2\). The electric field \(E\) due to a point charge \(q\) at distance \(r\) is given by \(E = (k*q) / r^2\), where \(k = 8.99 * 10^9 Nm^2/C^2\) is Coulomb's constant. Therefore, \(E1 = - (8.99 * 10^9 * -5.00 * 10^-9) / (0.04)^2 = 28.0875 N/C\) and \(E2 = (8.99 * 10^9 * 3 * 10^-9) / (0.05)^2 = 32.388 N/C\). The negative sign for \(E1\) denotes the electric field directed towards the negative charge.
04

Identify the Directions

The electric field due to \(q1\) at \(P\) points towards \(O\), so its unit vector \(\hat{r1} = -\hat{j}\). Similarly, the electric field due to \(q2\) at \(P\) points away from \(A\), so its unit vector \(\hat{r2} = (\cos \theta \hat{i} + \sin \theta \hat{j})\), where \( \theta = \arctan{0.04/0.03}\). Then \(\hat{r2} = 0.6 \hat{i} + 0.8 \hat{j}\). Thus, express the electric fields \(E1\) and \(E2\) in unit vectors as \(E1 = 28.0875 N/C * -\hat{j} = -28.0875 \hat{j} N/C\) and \(E2 = 32.388 N/C * (0.6 i + 0.8j) = 19.4328 \hat{i} + 25.9104 \hat{j} N/C\).
05

Calculate the Resultant Electric Field

The total electric field at point \(P\), \(E_{tot}\), is a vector sum of \(E1\) and \(E2\). Hence, \(E_{tot} = E1 + E2 = (-28.0875 \hat{j} + 19.4328 \hat{i} + 25.9104 \hat{j}) = 19.4328 \hat{i} - 2.1771 \hat{j} N/C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is fundamental in calculating the electric forces between charged objects. It states that the electric force between two point charges is directly proportional to the product of their charges. It is inversely proportional to the square of the distance separating them.
The defining equation is:
  • \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
Here, \( F \) represents the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \mathrm{Nm^2/C^2} \).
We also apply this principle when calculating electric fields, where the electric field \( E \) at a distance \( r \) from a charge \( q \) is given by:
  • \[ E = \frac{k \cdot q}{r^2} \]
Calculating electric fields from charges is essential to predict how charged objects interact at a distance.
Vector Addition
In physics, vector addition is crucial for understanding how different vectors combine. Electric fields, like many physical quantities, are vectors. They have both magnitude and direction.
When calculating the resultant electric field at a point due to multiple charges, these individual electric fields need to be added vectorially. This means considering both the directions and the magnitudes of the vectors:
  • If two vectors \( \vec{E_1} \) and \( \vec{E_2} \) are in the same or opposite directions, they can be easily added or subtracted by directly working with their magnitudes.
  • If they are at an angle to each other, decomposing each vector into components is necessary.
The resultant vector is found by summing the components along each axis. This approach enables us to find the net electric field resulting from multiple charges.
Pythagorean Theorem
The Pythagorean theorem is a key mathematical concept in physics, particularly in scenarios involving perpendicular distances and right triangles. It states that in a right-angled triangle:
  • \[ c^2 = a^2 + b^2 \]
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides.
This theorem helps when determining distances in a two-dimensional plane. In the given exercise, it was used to calculate the distance between charges and a point in the coordinate system by treating each spatial position as part of a right triangle.
Determining these distances accurately is vital for applying Coulomb's Law, as the strength of the electric field depends on these distances.
Coordinate System in Physics
Understanding the coordinate system is important in physics to specify positions and directions. A coordinate system allows us to precisely describe where objects are located and how they move.
In the exercise, we used a two-dimensional Cartesian coordinate system, which uses the x-axis and y-axis. Each point or object, like a charge or a point where the electric field is measured, can be plotted using these axes.
Different positions are represented as coordinates \( (x, y) \). This makes calculations, such as those involving distances or angles, more straightforward. By converting positions into a standard coordinate system, we can apply mathematical tools like vector addition and the Pythagorean theorem more effectively.

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Most popular questions from this chapter

An American penny is \(97.5 \%\) zinc and \(2.5 \%\) copper and has a mass of \(2.5 \mathrm{~g}\). (a) Use the approximation that a penny is pure zinc, which has an atomic mass of \(65.38 \mathrm{~g} / \mathrm{mol},\) to estimate the number of electrons in a penny. (Each zinc atom has 30 electrons.) (b) Estimate the net charge on all of the electrons in one penny. (c) The net positive charge on all of the protons in a penny has the same magnitude as the charge on the electrons. Estimate the force on either of two objects with this net magnitude of charge if the objects are separated by \(2 \mathrm{~cm}\). (d) Estimate the number of leaves on an oak tree that is 60 feet tall. (e) Imagine a forest filled with such trees, arranged in a square lattice, each \(10 \mathrm{~m}\) distant from its neighbors. Estimate how large such a forest would need to be to include as many leaves as there are electrons in one penny. (f) How does that area compare to the surface area of the earth?

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)

(a) What must the charge (sign and magnitude) of a \(1.45 \mathrm{~g}\) particle be for it to remain stationary when placed in a downwarddirected electric field of magnitude \(650 \mathrm{~N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(3.33 \times 10^{-21} \mathrm{~N} ?\)

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{C}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m}\) (d) \(x=0, y=0.200 \mathrm{~m}\)

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