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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 \(\hat{\jmath}\)

Short Answer

Expert verified
The unit vector at point (a) is -\(\hat{\jmath}\), at point (b) is 0.71\(\hat{\imath}\) + 0.71\(\hat{\jmath}\) and at point (c) is -0.39\(\hat{\imath}\) + 0.93\(\hat{\jmath}\)

Step by step solution

01

Solving for \(\hat{r}\) - Point A

To solve the unit vector in the direction of point (0,-1.35m), treat the origin as the base point (0,0). We calculate the magnitude of the vector \(r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). So here our calculation breaks down to \(r= \sqrt{(0-0)^2 + (-1.35-0)^2}\). So, \(r=1.35m\). Now the vector that points from the origin to the location is simply (-1.35)\(\hat{\jmath}\), and, normalizing this, we get the unit vector \(\hat{r}\) = (-1.35/1.35)\(\hat{\jmath}\) = -\(\hat{\jmath}\)
02

Solving for \(\hat{r}\) - Point B

As before, treat the origin as the base point (0,0) and point B as (0.12m , 0.12m). The magnitude of vector, \(r= \sqrt{(0.12-0)^2 + (0.12-0)^2}\). So, we have \(r= \sqrt{(0.12)^2 *2}\), which simplifies to \(r= 0.17m\). The vector that points from the origin to the location is (0.12)\(\hat{\imath}\) + (0.12)\(\hat{\jmath}\), normalizing this, we get the unit vector \(\hat{r}\) = [(0.12/0.17)\(\hat{\imath}\) + (0.12/0.17)\(\hat{\jmath}\)] which simplifies to 0.71\(\hat{\imath}\) + 0.71\(\hat{\jmath}\)
03

Solving for \(\hat{r}\) - Point C

Keeping the same approach, the origin as the base point (0,0) and point C as (-1.1m , 2.6m). The magnitude of vector, \(r= \sqrt{(-1.1-0)^2 + (2.6-0)^2}\). Hence, \(r= \sqrt{(1.1)^2 + (2.6)^2}\) which simplifies to \(r= 2.8m\). The vector from origin to the point is (-1.1)\(\hat{\imath}\) + (2.6)\(\hat{\jmath}\), normalizing this, we get the unit vector \(\hat{r}\) = [(-1.1/2.8)\(\hat{\imath}\) + (2.6/2.8)\(\hat{\jmath}\)] which simplifies to -0.39\(\hat{\imath}\) + 0.93\(\hat{\jmath}\)

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Key Concepts

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

Vector Calculations
In physics, especially in electrostatics, vector calculations are fundamental when dealing with forces and fields. Vectors are quantities characterized by both a magnitude and a direction. To calculate vectors, one must consider both of these elements.

When determining vectors from one point to another, think of the change in position as a straight line connecting the two points. One calculates the vector by finding the difference in the coordinates of the points along each dimension, whether it's in 2D or 3D space. For example, for points A and B, the vector is calculated using the formula \( \vec{r} = (x_2 - x_1)\hat{\imath} + (y_2 - y_1)\hat{\jmath} \).
  • The change along the x-axis is \( x_2 - x_1 \)
  • The change along the y-axis is \( y_2 - y_1 \)
This approach is useful to visualize how one point moves relative to another.

Once the vector is determined, its magnitude or length can be calculated using the Pythagorean theorem: \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). This magnitude provides us with the size of the vector but not the specific direction. To specify direction, we need to use unit vectors.
Unit Vectors
Unit vectors are crucial components in vector calculations as they express the direction of a vector without regard to its magnitude. They provide a universal way of representing direction in space, allowing for clarity and consistency in calculations.

A unit vector has a magnitude of one and indicates direction precisely. It is the normalized form of any given vector, calculated by dividing each component of the vector by its magnitude. This normalization process ensures that the resulting vector has the same direction as the original but with a magnitude of one.

Consider a vector \( \vec{r} = a\hat{\imath} + b\hat{\jmath} \); the unit vector \( \hat{r} \) is determined using: \[ \hat{r} = \frac{a}{r}\hat{\imath} + \frac{b}{r}\hat{\jmath} \] where \( r = \sqrt{a^2 + b^2} \). This normalization ensures we understand the direction explicitly while abstracting away its magnitude.
  • Example: A vector 3 units long in the positive x-direction has a unit vector \( \hat{\imath} \) in the same direction.
  • If it were directed 2 units into the negative y-direction, its unit vector would be \( -\hat{\jmath} \).
Unit vectors are particularly useful in expressing electric fields' directions without getting bogged down by the field's strength.
Electric Field Direction
Understanding the direction of electric fields is essential in electrostatics, as it reveals how charged objects influence their surroundings. The electric field direction at a point indicates the force's direction that a positive test charge would experience at that location.

Electric fields are vector fields, meaning they have both magnitude and direction. The field's direction is commonly established once its vector representation is known. This is where the concept of unit vectors becomes useful, as they enable us to identify direction independent of strength.

Calculating the electric field direction involves determining the unit vector from the source point (charge) to the field point (where direction is calculated). The unit vector gives the field's direction, while the magnitude tells us about the force's intensity.
  • For instance, given a charge at the origin, the electric field at a point \((x, y)\) is directed along the unit vector \( \hat{r} \) pointing from the origin to that point.
  • The electric field lines will always point away from positive charges and towards negative charges.
This visualization aids in understanding interactions between charges and predicting how changes in position alter the field's direction. Recognizing these directional cues is key to solving complex electrostatic problems.

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Most popular questions from this chapter

A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately \(10.0 \mu \mathrm{m}\) in diameter, and measurements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e\) ) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a \(0.10 \mathrm{~mm}\) length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a \(0.10 \mathrm{~mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is \(5.00 \mathrm{~cm}\) below the skin? (c) Certain shark can respond to electric fields as weak as \(1.0 \mu \mathrm{N} / \mathrm{C}\). How far from this segment of axon could a shark be and still detect its electric field?

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(1.60 \mathrm{~cm}\) distant from the first, in a time interval of \(3.20 \times 10^{-6} \mathrm{~s} .\) (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Positive charge \(Q\) is distributed uniformly along the positive \(y\) -axis between \(y=0\) and \(y=a\) A negative point charge \(-q\) lies on the positive \(x\) -axis, a distance \(x\) from the origin (Fig. P21.80). (a) Calculate the \(x\) - and \(y\) -components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\) -axis. (b) Calculate the \(x\) - and \(y\) -components of the force that the charge distribution \(Q\) exerts on \(q\) (c) Show that if \(x \gg a, F_{x} \cong-Q q / 4 \pi \epsilon_{0} x^{2}\) and \(F_{y} \cong+Q q a / 8 \pi \epsilon_{0} x^{3} .\) Explain why this result is obtained.

A charge \(+Q\) is located at the origin, and a charge \(+4 Q\) is at distance \(d\) away on the \(x\) -axis. Where should a third charge, \(q\), be placed, and what should be its sign and magnitude, so that all three charges will be in equilibrium?

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m} ?\)

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