Concept explainers
(a)
The electric field at x = 5.00 cm.

Answer to Problem 16PE
The electric field at x = 5.00 cm is
Explanation of Solution
Given:
Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.
Formula used:
Electric field in point P distanced di from n charges on a line is given with
Where positive sign is taken for charges left to the point P and negative for charges right to P.
Calculation:
Conclusion:
The electric field at x = 5.00 cm is
(b)
To Find:
The position between 3.00 and 8.00 cm where the total electric field the same as that for − 2q alone.

Answer to Problem 16PE
The position is 7 cm.
Explanation of Solution
Given:
Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.
Calculation:
The charges will cancel out at the mid-point of
So, we get
Conclusion:
Thus, the position is at 7 cm.
(c)
Whether the electric field between 0.00 and 8.00 cmbe zero.

Answer to Problem 16PE
Electric filed will never be zero.
Explanation of Solution
Given:
Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.
Calculation:
Note that such cancellation can't happen between a
which is not possible.
In other words, when it reaches 0 after continuously dropping in some direction, electric field will change direction.
Since
Which means it points to the left, so E = 0 can't happen neither it can happen between
Conclusion:
Electric field will never be zero.
(d)
To Find: where the electric field most rapidly approaches zero for very large positive or negative values of x in case (a) and (b).

Answer to Problem 16PE
Electric field of configuration (a) drops more rapidly.
Explanation of Solution
Given:
Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.
Calculation:
Total charge in (a) is zero, while total charge in (b) is not zero.
Looked from infinity, those charges may be considered emerging from the same point, so electric field in (a) will drop more rapidly.
Conclusion:
Thus, Electric field of configuration (a) drops more rapidly
(e)
Find the position to the right of 11.00 cm where the total electric field is zero, other than at infinity.

Answer to Problem 16PE
The position to the right of 11.00 cm where the electric field is zero other than infinity is 30.872 cm.
Explanation of Solution
Given info:
Point charges located at 3.00, 8.00 and 11.0 cm along the x -axis.
Calculation:
After cancelling some constants, we can obtain an equation,
This after some algebraic manipulation converts to,
With one real solution, x = 19.872 cm. since x denotes distance from the rightmost charge, coordinate of such point in the system given is,
Conclusion:
Thus, the position to the right of 11.00 cm where the electric field is zero other than infinity is 30.872 cm.
Want to see more full solutions like this?
Chapter 18 Solutions
COLLEGE PHYSICS
- Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µCarrow_forwardTwo conductors having net charges of +14.0 µC and -14.0 µC have a potential difference of 14.0 V between them. (a) Determine the capacitance of the system. F (b) What is the potential difference between the two conductors if the charges on each are increased to +196.0 µC and -196.0 µC? Varrow_forwardPlease see the attached image and answer the set of questions with proof.arrow_forward
- How, Please type the whole transcript correctly using comma and periods as needed. I have uploaded the picture of a video on YouTube. Thanks,arrow_forwardA spectra is a graph that has amplitude on the Y-axis and frequency on the X-axis. A harmonic spectra simply draws a vertical line at each frequency that a harmonic would be produced. The height of the line indicates the amplitude at which that harmonic would be produced. If the Fo of a sound is 125 Hz, please sketch a spectra (amplitude on the Y axis, frequency on the X axis) of the harmonic series up to the 4th harmonic. Include actual values on Y and X axis.arrow_forwardSketch a sign wave depicting 3 seconds of wave activity for a 5 Hz tone.arrow_forward
- Sketch a sine wave depicting 3 seconds of wave activity for a 5 Hz tone.arrow_forwardThe drawing shows two long, straight wires that are suspended from the ceiling. The mass per unit length of each wire is 0.050 kg/m. Each of the four strings suspending the wires has a length of 1.2 m. When the wires carry identical currents in opposite directions, the angle between the strings holding the two wires is 20°. (a) Draw the free-body diagram showing the forces that act on the right wire with respect to the x axis. Account for each of the strings separately. (b) What is the current in each wire? 1.2 m 20° I -20° 1.2 marrow_forwardplease solve thisarrow_forward
- please solve everything in detailarrow_forward6). What is the magnitude of the potential difference across the 20-02 resistor? 10 Ω 11 V - -Imm 20 Ω 10 Ω 5.00 10 Ω a. 3.2 V b. 7.8 V C. 11 V d. 5.0 V e. 8.6 Varrow_forward2). How much energy is stored in the 50-μF capacitor when Va - V₁ = 22V? 25 µF b 25 µF 50 µFarrow_forward
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





