In Fig. 22-44, a thin glass rod forms a semicircle of radius r = 3.87 cm….
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Question “In Fig. 22-44, a thin glass rod forms a semicircle of radius r = 3.87 cm….”
In Fig. 22-44, a thin glass rod forms a semicircle of radius r
= 3.87 cm. Charge is uniformly distributed along the rod, with +q =
3.52 pC in the upper half and -q = -3.52 pC in the lower half. What
is the magnitude of the electric field at P, the center of the
semicircle?
Answer
Concepts and Reason
This problem can be solved by using an electric field in the middle of a circular arch.
Calculate the electric field for each quarter at the center.
To calculate the net electric fields, you will need to take the vector sum due to the two-quarters.
Then, add the electric field expression to the entire arc and enter the values to calculate its magnitude.
Fundamentals
This expression shows the electric field that is at the center due to charge distribution in a circular arc.
denotes the permissivity of free space.
indicates the amount of charge.
denotes the distance between the point and the charge.
denotes the angle.
The net electric field
is the center. It’s given by the vector sum of electric fields due to two quarters:
The electric field is formed by
and
.
Locate the electric field in the middle of the circular arc caused by the upper half.
Consider a line element with angular width HTMLmedia_tag_9$making angle HTMLmedia_tag_10$with a positive y-axis and uniform linear charge density .
Let Q be charge at center of circle r.
The small line element produces an electric field that is called,
The following linear charge density formula is used:
The electric field caused by small elements is therefore,
The horizontal component points in the positive x direction, while the vertical component points towards negative y- direction.
Below is an illustration of electric field in vector form:
The electric field due to the lower arc will be given below, as the upper and lower halves of the arcs are symmetrical with respect to horizontal.
The vector sum of these two fields is the net electric field.
Locate the electric field in the middle of the semicircle.
The net electric field in the center is due to small elements.
Total electric field is due to the whole arc.
Substitute
to
,
to q,
to
,
or
, and
for r:
As follows:
The electric field acts in a negative y direction. The magnitude of the electric force is $media_tag_30.
Ans:
The electric field in the semi-circle’s center is .
Conclusion
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