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Figure 22-39 shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r=2.00 cm, with angles,,, and. What are the (a) magnitude and (b) direction (relative to the positive direction of the xaxis) of the net electric field produced at the center of the arc?

Short Answer

Expert verified
  1. The magnitude of the net electric field produced at the center of the arc is 3.92×10-6N/C.
  2. The direction of the net electric field produced at the center of the arc is data-custom-editor="chemistry" -76.4.

Step by step solution

01

The given data

  1. Radius of the circular disc,r=2.00cm
  2. Arrangement of electrons and protons at angles θ1=30.0, θ2=50.0, θ3=30.0, and θ4=20.0.
02

Understanding the concept of electric field

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again using the superposition law, we can get the value of the electric field in its direction and this determines the net electric field at that point.

The magnitude of the electric field is given as-

E=q4πεoR (i)

where, R = The distance of the field point from the charge

q = charge of the particle,

data-custom-editor="chemistry" εois the permittivity of vacuum.

The magnitude of the resultant vector,

V=Vx2+Vy2 (ii)

where, Vx is the horizontal component of the vector and Vy is the vertical component of the vector.

The angle or the direction of the electric field made with the x-axis,

data-custom-editor="chemistry" θ=tan-1VyVx (iii)

where, Vx is the horizontal component of the vector and Vy is the vertical component of the vector.

03

a) Calculation of the magnitude of the net electric field.

As the magnitude of the charge is the same on all the given particles and all the charges are at the same distance from the center, the magnitude of electric fields at the center will also be the same. The field of each charge can be given using equation (i) and the given values, as:

The directions of the electric fields due to the charge particles are indicated in the phasor diagram below.


Now, the components of electric field in both the x- and y-directions due to the produced fields can be given as: (for clockwise directions)

x-component

y-component

EA

Ecos0

Esin0

EB

Ecos-150o

Esin-150o

EC

Ecos-100o

Esin-100o

ED

Ecos130o

Esin130o

EE

Ecos-20o

Esin-20o

Thus, the net electric field along x-axis can be given as the sum of all the x-directional electric fields as follows:

Ex,net=Ecos0o+Ecos-150o+Ecos-100o+Ecos130o+Ecos-20o=3.6×10-9N/Ccos0o+cos-150o+cos-100o+cos130o+cos-20o=9.26×10-7N/C

Now, the net electric field along y--axis can be given as the sum of all the x-directional electric fields as follows:

Ey,net=Esin0o+Esin-150o+Esin-100o+Esin130o+Esin-20o=3.6×10-9N/Csin0o+sin-150o+sin-100o+sin130o+sin-20o=-3.81×10-6N/C

Thus, the magnitude of the resultant electric field is given using equation as follows:

localid="1662519095157" Enet=Ex,net2+Ey,net2=9.26×10-7N/C2+-3.81×10-6N/C2=3.92×10-6N/C

Hence, the magnitude of the net electric field is3.92×10-6N/C

04

b) Calculation of the direction of the net electric field

Now, the direction of the net electric field using the data obtained in part (a) calculations can be given using equation (iii) as:

θ=tan-1-3.81×10-6N/C9.26×10-7N/C=-76.4o

Hence, the direction of the field is -76.4ofrom the positive of x-axis.

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