To determine Electric force on the charge. Explanation Consider two rain drops of equal charge q on x-axis which are 2.18 mm apart and a rain drop with charge Q 12.3 mm up on the y-axis. Write the equation for net force on upper rain drop. F Q = F Q q + F Q q = 2 F Q q (I) Here F Q is the force on upper charge and F Q q is the force on charge Q due to q . Expand (I) F Q → = 2 ( K Q q r 2 ) r ^ (II) Here K is the proportionality constant, Q and q are the charges, r is the distance of separation between the charges and r ^ is the unit vector that points from Q to q . Use Pythagoras theorem to expand r 2 . r 2 = ( x 2 ) 2 + y 2 (III) Write the equation for unit vector r ^ r ^ = y ( x 2 ) 2 + y 2 (IV) Substitute (III) and (IV) in (II) F Q = 2 ( K Q q ( x 2 ) 2 + y 2 ) y ( x 2 ) 2 + y 2 (V) Conclusion: Substitute 9 × 10 9 N ⋅ m 2 / C 2 for K , 645 nC for q , − 1.87 μC for Q , 2.18 mm for x and 12.3 mm for y in (V) F Q = 2 ( ( 9 × 10 9 N ⋅ m 2 / C 2 ) ( − 1.87 μC ( 10 -6 C 1 μC ) ) ( 645 nC ( 10 -9 C 1 μC ) ) ( 2

4th Edition

ISBN: 9780134988566

Author: Wolfson, Richard

Publisher: Pearson Education,

See similar textbooks

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Videos

Question

Chapter 20, Problem 36E

To determine

Electric force on the charge.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 20, Problem 35E

Chapter 20, Problem 37E

Students have asked these similar questions

You are standing a distance x = 1.75 m away from this mirror. The object you are looking at is y = 0.29 m from the mirror. The angle of incidence is θ = 30°. What is the exact distance from you to the image?

For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

Chapter 20 Solutions

Essential University Physics

Ch. 20.1 - The proton is a composite particle composed of...Ch. 20.2 - Charge q1 is located at x = 1 m, y = 0. What...Ch. 20.3 - A positive point charge is located at the origin...Ch. 20.4 - Far from a charge distribution, you measure an...Ch. 20.5 - An electron, a proton, a deuteron (a neutron...Ch. 20 - Conceptual Example 20.1 shows that the...Ch. 20 - A free neutron is unstable and soon decays to...Ch. 20 - Where in Fig. 20.5 could you put a third charge so...Ch. 20 - Equation 20.3 gives the electric field of a point...Ch. 20 - Is the electric force on a charged particle always...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.