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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

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

(a) An electron is moving east in a uniform electric field of \(1.50 \mathrm{~N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{~m}\) east of point \(A ?\) (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{~m} / \mathrm{s},\) east. What is the speed of the proton at point \(B ?\)

(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?

Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.

If two electrons are each \(1.50 \times 10^{-10} \mathrm{~m}\) from a proton (Fig. E21.41), find the magnitude and direction of the net electric force they will exert on the proton.

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