Question

In the figure below, a nonconducting rod of length L = 7.50 cm has charge -q = -4.36 fC uniformly distributed along its length.
Note= f is femto 10^-15

(a) What is the linear charge density of the rod?
____ C/m

(b) What is the magnitude of the electric field at point P, a distance a = 12.0 cm from the end of the rod?
____ N/C

(c) What is its direction?
______° (counterclockwise from the positive x axis)

(d) What is the electric field magnitude produced at distance a = 50 m by the rod?
______ N/C

(e) Repeat part (d) for a particle of charge -q = -4.36 fC that replaces the rod.
______ N/C

Answer 1

Concepts and reason

The concepts used to solve this problem are linear charge density, and electric field due to point charge.

Use the relation between linear charge density, charge and length of rod to calculate the linear charge density of the rod.

Use the concept of electric field due to point charge to calculate magnitude of electric field at certain point.

Use the sign of electric field to determine its direction.

Fundamentals

The charge per unit length is known as linear charge density.

Expression for linear charge density is,

Here, the charge density is , charge is , and the length is .

Expression for electric field due to a point charge is,

$dE=kdQ $

Here, the electric field is , coulomb’s constant is , charge is , and the distance is .

(a)

The following figure depicts the non conducting rod.

The ratio between the total charge and the length is linear charge density.

Expression for linear charge density is,

Substitute for and for.

The linear charge density of the non-conducting rod is .

(b)

The electric field magnitude is equal to the product of coulomb’s constant with the charge divided by the square of the distance.

Expression for electric field due to point charge is,

$dE=kdQ $

The expression for the unit charge is,

The electric field of component is,

The total electric field produced at point is,

Substitute for.

Solve for

Substitute for, for , for , and for.

The magnitude of electric field is,

(c)

The electric field has a negative sign. This means, electric field is travelling in negative

direction.

The negative axis is at with the positive axis.

Since the direction of electric field is towards negative axis which is with the positive axis in counterclockwise direction.

The direction of electric field is towards negative axis which is with the positive axis in counterclockwise direction.

(d)

The given value of distance is which is very large as compared to the length of the rod.

Since the value of can be considered as due to.

The expression for the electric field becomes,

Substitute for , for and for.

The magnitude of the electric field is,

(e)

The point charge replaces the rod, which means the distance is .

Since, expression for electric field becomes,

Substitute for, for , and for .

The magnitude of the electric field is,

.

Ans: Part a

The linear charge density of the non-conducting rod is .

Part b

The magnitude of the electric field at point P at a distance from end of the rod is .

Part c

The direction of the electric field is towards negative x axis which is with the positive x axis in counterclockwise direction.

Part d

The magnitude of the electric field produced at a distance by the rod is .

Part e

The magnitude of the electric field produced at a distance when the rod replaced is.