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A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{m} )\) and point \(b\) (at \(x=0.90 \mathrm{m} )\) is 240 \(\mathrm{V}\) . (a) Which point, \(a\) or \(b,\) is at the higher potential? (b) Calculate the value of \(E.\) (c) A negative point charge \(q=-0.200 \mu C\) is moved from \(b\) to \(a\) . Calculate the work done on the point charge by the electric field.

Short Answer

Expert verified
(a) Point \(a\) is at higher potential. (b) \( E = 800 \) V/m. (c) Work done is 48 \( \mu \)J.

Step by step solution

01

Understanding Electric Potential Difference

The potential difference between two points is given by \( V_b - V_a = -E \cdot d \), where \( d \) is the distance between the points and \( E \) is the electric field. Here, the potential difference \( V_a - V_b = 240 \) V is given, which means \( V_a > V_b \), as a positive potential difference indicates point \(a\) is at a higher potential.
02

Calculate the Electric Field Strength

From the potential difference equation \( V_a - V_b = -E \cdot d \), rearrange to find \( E = \frac{V_a - V_b}{-d} \). Substituting values, \( E = \frac{240}{-(0.90 - 0.60)} = \frac{240}{-0.30} = -800 \) V/m. The negative sign confirms the field is in the negative \(x\)-direction.
03

Calculate Work Done by the Electric Field

Work done, \( W \), by the electric field on a charge moving between two points is given by \( W = q(V_b - V_a) \). Substituting \( q = -0.200 \times 10^{-6} \) C and \( V_b - V_a = -240 \) V, we have \( W = (-0.200 \times 10^{-6}) \times (-240) = 48 \times 10^{-6} \) J or \( 48 \mu \)J.

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Key Concepts

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

Potential Difference
Potential difference is the measure of the energy difference per unit charge between two points in an electric field. It tells us how much work is needed to move a charge between these points. In the context of this problem, the potential difference between points \( a \) and \( b \) is given as 240 \( \mathrm{V} \). This indicates that point \( a \) is at a higher potential than point \( b \), since the potential difference is positive when calculated from \( a \) to \( b \).
For example, if you imagine a battery where one side is higher than the other, the side with more energy per electron is at a higher potential.
This concept is crucial because it allows us to calculate other aspects of the electric field like strength and work done.
Electric Potential
Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point, without any acceleration. It's a scalar quantity measured in volts (\( \mathrm{V} \)).
In a uniform electric field, the potential decreases linearly with distance in the field's direction. This means the farther you move against the field lines, the more work you have to put in, making the potential higher.
  • Higher potential: Point where you have to push more against the electric field.
  • Lower potential: Point where it becomes easier, requiring less push or work against the field.
This understanding helps connect the potential difference and electric field strength.
Work Done by Electric Field
The work done by an electric field when moving a charge from one point to another is calculated using the formula \( W = q(V_b - V_a) \). It is essentially the energy required to move a charge between two points in an electric field.
In this problem, moving a negative point charge of \( q = -0.200 \mu \mathrm{C} \) from point \( b \) to point \( a \) implies the electric field is doing work on the charge. Substitution of values shows the work done is \( 48 \mu \mathrm{J} \).
It is important to remember:
  • Work done is positive if the field assists in moving the charge.
  • Work done is negative if you have to do the work against the field.
This concept provides insight into how energy is transferred in an electric field.
Uniform Electric Field
A uniform electric field has a constant magnitude and direction. When depicted in a diagram, the field lines are parallel and evenly spaced throughout the region.
In this problem, the field is directed in the negative \( x \)-direction. This setup means the same amount of force and energy change are experienced at any point in the field when moving a charge.
  • Fields like these simplify calculations since field strength and direction don't vary.
  • Potential difference across any two points is directly proportional to the distance between them.
Understanding a uniform electric field helps predict and calculate how charges behave within this field smoothly.
Point Charge
A point charge is a hypothetical charge located at a singular point in space. It makes calculations simpler by assuming the charge's size or shape doesn't affect the overall analysis.
This type of charge is pivotal in electrostatics since it allows us to conceptualize how electric fields interact with a charged particle.
  • For our problem, a negative point charge \( -0.200 \mu \mathrm{C} \) is moved, highlighting how it experiences the electric field.
  • Makes it easier to calculate forces, work done, and potential changes due to its independent nature.
Understanding the characteristics of a point charge enhances our ability to model and predict physical phenomena in electric fields.

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Most popular questions from this chapter

Deffecting Plates of an Oscilloscope. The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 \(\mathrm{cm}\) on a side, with a separation of about 5.0 \(\mathrm{mm} .\) The potential difference between the plates is 25.0 \(\mathrm{V} .\) The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 \(\mathrm{cm}\) apart (Fig. \(\mathrm{P} 23.62 ) .\) The plates are insulating and have uniform surface charge densities \(+\sigma\) and \(-\sigma .\) The charge on the sphere is \(q=8.90 \times 10^{-6} \mathrm{C} .\) What potential difference between the plates will cause the thread to assume an angle of \(30.0^{\circ}\) with the vertical?

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J.}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of \(\vec{\boldsymbol{E}}\)?

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a,\) and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{J} .\) When the second charge is moved to point \(b,\) the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

(a) An electron is to be accelerated from 3.00 \(\times 10^{6} \mathrm{m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

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