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In Fig. 22-50 positive charge q = 8.30 pC is spread uniformly along a thin nonconducting...

In Fig. 22-50 positive charge q = 8.30 pC is spread uniformly along a thin nonconducting rod of length L = 17.0 cm. What are the (a)x- and (b)y- components of the electric field (in N/C) produced at point P, at distance R = 6.00 cm from the rod along its perpendicular bisector?
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Concepts and reason

The concept required to solve this problem is electric field.

Initially, draw the diagram for the electric field at the field point due to any two symmetric charge elements. Then, write the expression for the y- component of electric field due to an infinitely small charge element.

After that substitute, the expressions for small charge and the distance between the field point and chare element and integrate for the for the electric field due to complete rod.

For the numerical value substitute the values in the expression and solve.

Fundamentals

The expression of the electric field is,

Here, E is the electric field, k is the coulomb’s constant, Q is the charge, and R is the distance.

The expression of the linear charge density is,

Here, is the linear charge density, Q is the charge, and L is the length.

(a)

Consider a uniformly charged rod. The electric field at a point on the perpendicular bisector at distance from the rod is shown in the following diagram.

Ē,
Esine
À EN
Esine
Ē, cose
Ecos e

Here, is the angle made by the line joining charge element to test point with the horizontal, is the length of the rod and is the distance between the charge element and test point.

and are electric field at the field point due to two mall symmetric elements on the two sides of the bisector.

From the given diagram, due to symmetry the horizontal component of electric field cancels out each other.

(b)

The expression of the linear charge density is,

The charge due to the small segment in terms of is,

dQ = Adx

The component of electric field due to small charge is,

dE, = kall sino

Substitute for , (x + R2)
for and (x + R22
for as follows:de __
k(Adr)
R
{(x?+R2)??* (x + R2)?
_kadxR
*(x? + R2)72

So, the net electric field at the field point along direction is,

E, =2. kader
• (x’ + R2)92
di
= 2kar
? (x² + R2)52

Substitute for and solve for the as follows:

5,-3? * *
2k1
| R I +4R?

The net electric field at point field point is,

RL +4R

Substitute 8.30 pC
for , 17.0 cm
for , 6.00 cm
for and 8.987x10°N-m²/C
for as follows:

2(8.30 pC)(1x10-12 C/PC)(8.987x10°N-m²/C)
(6.00 cm)(0.01 m/cm)
E =
/{(17.0 cm)+4(6.00 cm)}(0.01 m/cm)
=11.9 N/C

Ans:

The x- component of the electric field is ON/C
.

The y- component of the electric field is 11.9 N/C
.

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