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A particle with charge +1nC (a nanocoulomb is 1X10-9C) is located at the origin. What is the electric field due to this particle at a location <0.1,0,0>m?

Short Answer

Expert verified

The electric field due to this particle at a location 0.1,0,0m is 9.0×102N/C .

Step by step solution

01

Identification of the given data

The given data can be listed below as:

  • The electric charge on a particle is, q=1×10-9C.
  • The location of the charged particle is, x1,y1,z1=0,0,0m.
  • The location of another point is, x2,y2,z2=0.1,0,0m.
02

Significance of electric field

Due to a specific point charge, the electric field exists in a spherical area surrounding the point charge. The value of the electric field depends inversely on the distance from the specific point charge.

03

Determination of the position vector from the origin to the specified location

The expression of the position vector from the origin to the specified location can be expressed as:

r=x2,y2,z2-x1,y1,z1

Here, rrepresents the position vector from the origin to the specified location.

Substitute all the values in the above equation.

r=0.1,0,0m-0,0,0m=0.1,0,0m

04

Determination of the magnitude of the distance from the origin to the specified location

The magnitude of the distance from the origin to the specified location can be calculated as:

r=r=0.1m2+om2+0m2=0.1m

05

Determination of the electric field due to this particle at a location <0.1,0,0>m

The expression of the electric field due to this particle at a location can be expressed as:

E=kqr2

Here, E represents the required electric field due to this particle at a location 0.1,0,0mand k represents the Coulomb’s constant whose value is 9×109N.m2/C2.

Substitute the values in the above equation.

E=9×109N.m2/C21×10-9C0.1m2=9.0×102N/C

Hence, the electric field due to this particle at a location 0.1,0,0mis 9.0×102N/C.

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Most popular questions from this chapter

In Figure 13.66 a proton at location A makes an electric field E1at location B. A different proton, placed at location B, experiences a force F1. Now the proton at B is removed and replaced by a lithium nucleus, containing three protons and four neutrons. (a) Now what is the value of the electric field at location B due to the proton? (b) What is the force on the lithium nucleus? (c) The lithium nucleus is removed, and an electron is placed at location B. Now what is the value of the electric field at location B due to the proton? (d) What is the magnitude of the force on the electron? (e) Which arrow in Figure 13.65 best indicates the direction of the force on the electron due to the electric field?

A dipole is located at the origin and is composed of charged particles with charge +eand-e , separated by a distance6×10-6m along the y axis. The +echarge on -y axis. Calculate the force on a proton due to this dipole at a location(0,4×10-8,0)m .

A charge of +1 nC1×109 Cand a dipole with charges +qand -qseparated by 0.3 mmcontribute a net field at location A that is zero, as shown in Figure 13.74.

(a) Which end of the dipole is positively charged? (b) How large is the charge?

Two dipoles are oriented as shown in Figure 13.72. Each dipole consists of two charges +qand -q, held apart by a rod of length s, and the center of each dipole is a distance dfrom location A. If=2nC, s=1mmand d=8cm, what is the electric field at location A? (Hint: Draw a diagram and show the direction of each dipole’s contribution to the electric field on the diagram.)

The dipole moment of the HF (hydrogen fluoride) molecule has been measured to 6.3×10-30Cm.. If we model the dipole as having charges of +eand -e separated by a distance s, what is s ? Is this plausible?

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