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A point charge q1=+2.40 μC is held stationary at the origin. A second point charge q2=4.30 μC moves from the point x= 0.150 m, y= 0 to the point x= 0.250 m, y= 0.250 m. How much work is done by the electric force on q2?

Short Answer

Expert verified
The work done by the electric force is approximately -3.56 Joules.

Step by step solution

01

Understand the Problem

We need to calculate the work done by the electric force as charge q2 moves from one point to another in the electric field of charge q1. The two charges have values of q1=+2.40μC and q2=4.30μC.
02

Initial and Final Positions

The initial position of q2 is (x1,y1)=(0.150,0) and the final position is (x2,y2)=(0.250,0.250).
03

Calculate Initial Distance

Calculate the distance r1 between q1 and the initial position of q2. Since q1 is at the origin, r1=(0.150)2+02=0.150 m.
04

Calculate Final Distance

Calculate the distance r2 between q1 and the final position of q2. Since q2 moves to point (0.250,0.250), r2=(0.250)2+(0.250)2=0.1250.354 m.
05

Use Formula for Work Done

The work done W is given by the formula: W=kq1q2(1r21r1) where k=8.99×109Nm2/C2.
06

Substitute Values and Calculate

Substitute the known values into the formula: W=8.99×109×2.40×106×(4.30×106)(10.35410.150) Calculate W.
07

Final Calculation

After calculating, W3.56Joules, indicating that the electric force does this amount of work on q2 while moving between the two points.

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Key Concepts

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

Point Charge
In the world of electromagnetism, a "point charge" refers to an idealized model of a charged object that is assumed to be infinitely small. This is helpful in simplifying calculations involving electric fields and forces.
This charge is treated as a single point in space, with no dimensions, allowing us to focus solely on the effects of the charge itself, rather than its size or shape. In practice, actual charges always occupy space, but this simplification helps physicists make useful predictions and calculations more efficiently.
For example, in the problem provided, both charges ( - one at the origin, - the other moving through space), can be considered point charges. This allows us to calculate the force and work done by the electric field generated by these charges using straightforward mathematical formulas.
Work Done
"Work done" by an electric force refers to the energy transferred by this force when a charge moves in an electric field. When a charge moves from one point to another, the electric force can do work on the charge, which can either be positive or negative depending on the movement direction relative to the force.
In the example exercise, we initially identify two positions for our second charge ( - an initial position at point (0.150, 0) m - and a final position at (0.250, 0.250) m). The force exerted by the electric field does work on the moving charge as it transitions between these locations.
This work done ( - calculated using the initial and final distances between charges), provides insight into how the energy of the system changes during the charge's movement. In our solution, the calculation showed that the work done by the electric force was approximately -3.56 Joules.
Electric Field
The "electric field" is a region around a charged object where other charges experience a force. This concept is crucial for understanding how charges interact in space.
An electric field is depicted as a vector field, which means it has both a magnitude and a direction at every point in space. The direction of the field at any point is the direction of the force that a positive test charge would experience at that point.
In our example, - the electric field created by point charge 1 influences the motion of point charge 2. As charge 2 moves, it experiences a force due to the electric field, which results in work being done, as previously discussed. Understanding how electric fields interact with charges is fundamental to solving problems involving electrostatic forces and energies.
Coulomb's Law
"Coulomb's Law" is a fundamental principle used to calculate the electric force between two point charges. It states that the force (- either attractive or repulsive)between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The mathematical equation for Coulomb's Law is:F=k|q1q2|r2where:
  • F is the magnitude of the force between the charges,
  • k is Coulomb's constant (8.99×109Nm2/C2),
  • |q1| and |q2| are the magnitudes of the charges,
  • r is the distance between the charges.
In the widget exercise, Coulomb's Law helps us calculate the initial and final forces, thus allowing us to determine the work done as one charge moves in the electric field created by another charge.

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Most popular questions from this chapter

Charge Q= 5.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R= 12.0 cm. A small sphere with charge q=+3.00 μC and mass 6.00 ×105 kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?

A hollow, thin-walled insulating cylinder of radius R and length L (like the cardboard tube in a roll of toilet paper) has charge Q uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if LR, the result of part (a) reduces to the potential on the axis of a ring of charge of radius R. (See Example 23.11 in Section 23.3.) (c) Use the result of part (a) to find the electric field at all points along the axis of the tube.

A positive point charge q1=+5.00×104 C is held at a fixed position. A small object with mass 4.00 ×103 kg and charge q2=3.00×104 C is projected directly at q1 . Ignore gravity. When q2 is 0.400 m away, its speed is 800 m/s. What is its speed when it is 0.200 m from q1 ?

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 m) and point b (at x= 0.90 m) is 240 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 μC is moved from b to a. Calculate the work done on the point charge by the electric field.

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by V(x)=Cx4/3 where x is the distance from the cathode and C is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of C. (b) Obtain a formula for the electric field between the electrodes as a function of x. (c) Determine the force on an electron when the electron is halfway between the electrodes.

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