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Chapter 18: Problem 35

A uniform electric field exists everywhere in the \(x, y\) plane. This electric field has a magnitude of \(4500 \mathrm{N} / \mathrm{C}\) and is directed in the positive \(x\) direction. A point charge \(-8.0 \times 10^{-9} \mathrm{C}\) is placed at the origin. Determine the magnitude of the net electric field at (a) \(x=-0.15 \mathrm{m}\) (b) \(x=+0.15 \mathrm{m}\) and \((\mathrm{c}) \mathrm{y}=+0.15 \mathrm{m}\)

Short Answer

Expert verified
Net fields: (a) 1300 N/C, (b) 7700 N/C, (c) 5480 N/C.

Step by step solution

01

Understand the Setup and Given Information

We are given a uniform electric field of magnitude \(E = 4500 \ \mathrm{N/C}\) directed along the positive \(x\)-axis. There is also a point charge \(q = -8.0 \times 10^{-9} \ \mathrm{C}\) located at the origin. We need to find the net electric field at different positions around this charge and field system.
02

Calculate the Electric Field Due to the Point Charge

The electric field from a point charge \(q\) at a distance \(r\) is given by: \[ E_q = \frac{k|q|}{r^2} \]Where \(k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}\). Since the charge is negative, the direction of the electric field will be towards the charge.
03

Find the Position (a) Net Electric Field at \(x = -0.15 \ \mathrm{m}\)

The distance \(r\) from the origin to \(x = -0.15 \ \mathrm{m}\) is 0.15 m. Thus, the electric field due to the charge is:\[ E_q = \frac{8.99 \times 10^9 \times 8.0 \times 10^{-9}}{(0.15)^2} \approx 3.20 \times 10^3 \ \mathrm{N/C} \]Since it's at a negative \(x\), the field from the charge points towards the origin. Net field: \(E_\text{net} = 4500 - 3200 = 1300 \ \mathrm{N/C} \) to the right.
04

Find the Position (b) Net Electric Field at \(x = +0.15 \ \mathrm{m}\)

The distance is the same, so the electric field magnitude from the charge is still roughly \(3200 \ \mathrm{N/C}\). Here, at positive \(x\), fields from both the uniform field and point charge point along \(x\), so they sum:\(E_\text{net} = 4500 + 3200 = 7700 \ \mathrm{N/C} \) to the right.
05

Find the Position (c) Net Electric Field at \(y = +0.15 \ \mathrm{m}\)

Now, the point is directly above the charge along the \(y\)-axis. The distance \(r\) is again 0.15 m, so the field due to charge is vertical:\[ E_q = \frac{8.99 \times 10^9 \times 8.0 \times 10^{-9}}{(0.15)^2} \approx 3200 \ \mathrm{N/C} \]The uniform field only acts in the \(x\)-direction, so the net field components remain different, resulting in a combined net field calculated as:\(E_\text{net} = \sqrt{4500^2 + 3200^2} = 5480 \ \mathrm{N/C} \).

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Key Concepts

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

Point Charge
A point charge is an idealized model of a charged object where the electricity is concentrated at a single point in space with no actual size or shape. The electric field caused by a point charge is radial, either pointing away (positive charge) or towards the charge (negative charge). The strength of the electric field at a distance from the charge can be calculated using the formula:\[ E = \frac{k|q|}{r^2} \]where:
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}\).
  • \(q\) is the charge value.
  • \(r\) is the distance from the point charge.
A negative charge will have its field lines pointing in towards the charge, which affects both the direction and sign of the net electric field when interacting with other fields.
Uniform Electric Field
A uniform electric field is consistent and has the same strength and direction everywhere within a particular space. This field is often represented by parallel lines in diagrams, indicating no variation in strength. A common presence in physics problems, uniform fields appear in various real-world applications, such as capacitors.
The magnitude of the electric field is given, in this case, as \(4500 \ \mathrm{N/C}\), pointing in the positive \(x\)-direction. This means the electric field exerts the same force on charges located in this space, helpful for calculating net effects when combined with other fields, such as the electric field from a point charge.
Net Electric Field
The net electric field is a vector sum of all individual electric fields present at a particular location. When different fields overlap, we need to account for both their magnitudes and directions. Here's how to approach finding the net electric field:
  • Add the magnitudes if the fields point in the same direction.
  • Subtract the magnitudes if they point in opposite directions.
  • Use vector addition if fields are perpendicular, applying the Pythagorean theorem.
In the original exercise, the uniform electric field interacts with the field produced by a point charge, requiring careful consideration of field directions to compute the net result at various points.
Electric Field Magnitude
The magnitude of an electric field describes its strength and is measured in newtons per coulomb (\(\mathrm{N/C}\)). In problems like the one above, understanding the magnitude helps in calculating forces exerted on a charge placed within the field.
For a point charge, the magnitude depends on the charge's value and the distance from it, expressed as:\[ E_q = \frac{k|q|}{r^2} \]For the uniform field, magnitude remains constant at \(4500 \ \mathrm{N/C}\). Together, these magnitudes allow for determining the resulting force or action on other charges placed in or moving through these fields.

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Most popular questions from this chapter

Review the important features of electric field lines discussed in Conceptual Example 12. Three point charges \((+q,+2 q,\) and \(-3 q)\) are at the corners of an equilateral triangle. Sketch in six electric field lines between the three charges.

Four identical metal spheres have charges of \(q_{\Lambda}=-8.0 \mu \mathrm{C}, q_{\mathrm{B}}=\) \(-2.0 \mu \mathrm{C}, q_{\mathrm{c}}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C}\) (a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C} ?\) (c) The final charge on each of the three separated spheres in part (b) is \(+3.0 \mu \mathrm{C} .\) How many electrons would have to be added to one of these spheres to make it electrically neutral?

An unstrained horizontal spring has a length of \(0.32 \mathrm{m}\) and a spring constant of \(220 \mathrm{N} / \mathrm{m}\). Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by \(0.020 \mathrm{m}\) relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of \(+5 q .\) Sphere \(\mathrm{B}\) carries a charge of \(-q .\) Sphere \(\mathrm{C}\) carries no net charge. Spheres \(A\) and \(B\) are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Last, sphere \(C\) is touched to sphere \(\mathrm{B}\) and separated from it. (a) How much charge ends up on sphere C? What is the total charge on the three spheres (b) before they are allowed to touch each other and (c) after they have touched?

A surface completely surrounds a \(+2.0 \times 10^{-6} \mathrm{C}\) charge. Find the electric flux through this surface when the surface is (a) a sphere with a radius of \(0.50 \mathrm{m},\) (b) a sphere with a radius of \(0.25 \mathrm{m},\) and \((\mathrm{c})\) a cube with edges that are \(0.25 \mathrm{m}\) long.
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