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Chapter 20: Problem 23

What are the strength and direction of the electric field \(2.0 \mathrm{cm}\) from a small glass bead that has been charged to \(+6.0 \mathrm{nC} ?\)

Short Answer

Expert verified
The strength of the electric field is \( E = 1.35 \times 10^6 \, N/C \) while the direction is radially outward away from the bead.

Step by step solution

01

Identify the Given Values

From the problem description, it can be discerned that: the charge \( Q = +6.0 \mathrm{nC} = 6.0 \times 10^{-9} \mathrm{C} \) and the distance \( r = 2.0 \mathrm{cm} = 2.0 \times 10^{-2} \mathrm{m} \). The positive direction shows that the charge is positive.
02

Compute for the Electric Field Strength

Using the electric field formula \( E = k_e Q / r^2 \), substitute \( k_e = 9.0 \times 10^9 \, Nm^2/C^2 \), \( Q = 6.0 \times 10^{-9} \mathrm{C} \), and \( r = 2.0 \times 10^{-2} \mathrm{m} \) into the equation to solve for \( E \).
03

Identify the Direction of the Electric Field

For a positive source charge (like in this case), the electric field lines move away from the charge. Therefore, the direction of the electric field at a point 2.0cm from the bead will be pointing radially outward away from the bead.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter. It comes in two types: positive and negative. Charges interact with each other through electric forces.

Here are a few key points about electric charge:
  • Like Charges Repel: Positive charges repel other positive charges. Similarly, negative charges repel negative charges.
  • Opposite Charges Attract: A positive charge attracts a negative charge.
  • Quantization: Electric charge is quantized, meaning it comes in discrete amounts, the smallest being the charge of an electron or proton.
In the exercise, a small glass bead was charged to a positive value of 6.0 nanocoulombs (nC). This positive charge creates an electric field around the bead, influencing other charges in its vicinity.
Electric Field Strength
The electric field strength at a point in space is a measure of the force that a charge would experience at that point.

The formula to calculate electric field (E) is:\[E = \frac{k_e \cdot Q}{r^2}\]where:
  • \(E\) is the electric field strength.
  • \(k_e\) is Coulomb's constant, approximately \(9.0 \times 10^9 \, \mathrm{Nm^2/C^2}\).
  • \(Q\) is the charge, which in this case is \(6.0 \times 10^{-9} \, \mathrm{C}\).
  • \(r\) is the distance from the charge, \(2.0 \times 10^{-2} \, \mathrm{m}\).
This enables us to determine how strong the field is around the charge. A higher field strength means a stronger force exerted on other charges.
Electric Field Direction
Understanding the direction of the electric field is crucial. Electric field lines tell us the direction a positive test charge would move if placed in the field.

Here's how it works for different charges:
  • Positive Source Charge: Electric field lines radiate outward. If a positive charge is creating the field, like the glass bead in the exercise, the field lines point away from it.
  • Negative Source Charge: Electric field lines point inward, towards the charge.
In our example, since the glass bead is positively charged, the electric field at a distance of 2.0 cm from it points outward. This understanding helps predict interactions with other charges nearby.

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Most popular questions from this chapter

When a honeybee flies through the air, it develops a charge of \(+17 \mathrm{pC} .\) How many electrons did it lose in the process of acquiring this charge?

A plastic rod is charged to \(-20 \mathrm{nC}\) by rubbing. a. Have electrons been added to the rod or protons removed? Explain. b. How many electrons have been added or protons removed?

II A parallel-plate capacitor is constructed of two square plates, size \(L \times L,\) separated by distance \(d .\) The plates are given charge \(\pm Q .\) Let's consider how the electric field changes if one of these variables is changed while the others are held constant. What is the ratio \(E_{\mathrm{f}} / E_{\mathrm{i}}\) of the final electric field strength \(E_{\mathrm{f}}\) to the initial electric field strength \(E_{\mathrm{i}}\) if: a. \(Q\) is doubled? b. \(L\) is doubled? c. \(d\) is doubled?

Object A, which has been charged to \(+10 \mathrm{nC},\) is at the origin. Object \(\mathrm{B},\) which has been charged to \(-20 \mathrm{nC},\) is at \((x, y)=(0.0 \mathrm{cm}, 2.0 \mathrm{cm}) .\) What are the magnitude and direction of the electric force on each object?

I Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of \(+12 \mathrm{pC} ;\) these are both very common values. In a thunderstorm, the electric field under a cloud can reach \(15,000 \mathrm{N} / \mathrm{C},\) directed upward. For a droplet exposed to this field, how do the magnitude and direction of the electric force compare to those of the weight force?
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