Question

A uniform line charge extends from x = - 2.6 cm to x = + 2.6 cm and has a linear charge density of = 5.5 nC/m.(a) Find the total charge.Find the electric field on the y axis at the following distances.(b) y = 4 cm(c) y = 12 cm

Answer 1

To find the total charge of the uniform line charge, multiply the linear charge density by the length of the line. To find the electric field at a certain distance on the y axis, use the formula for electric field due to a line charge.

Step-by-step explanation:

(a) To find the total charge of the line, we need to multiply the linear charge density by the length of the line:

Charge = (Linear charge density)*(Length)

Convert the length to meters:

Length = (2.6 cm + 2.6 cm) = 0.052 m

Total Charge = (5.5 nC/m)*(0.052 m) = 0.286 nC

(b) To find the electric field at y = 4 cm, we can use the formula:

Electric Field = (Linear charge density)/(2πε₀y)

Convert the linear charge density to C/m:

Charge density = 5.5 nC/m = 5.5 x 10-9 C/m

Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.04 m))

(c) To find the electric field at y = 12 cm, we can use the same formula:

Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.12 m))

Answer 2

Answer with Explanation:

We are given that

x=-2.6 cm to x=2.6 cm

Linear charge density=
`\lambda=5.5nC/m=5.5* 10^(-9) C/m`

`1nC=10^(-9) C`

Length of wire=
`2.6-(-2.6)=2.6+2.6=5.2cm`

Length of wire=
`(5.2)/(100)=0.052m`

1 m=100 cm

We know that

a.Linear charge density=
`(Q)/(L)`

Where Total charge =Q

Length=L

Total charge,Q=
`\lambda L`

Using the formula

Total charge,Q=
`5.5* 10^(-9)* 0.052=2.86* 10^(-10)C`

b.y=4 cm=
`(4)/(100)=0.04m`

1 m=100 cm

Electric field on the y-axis is given by

`E=(2\lambda)/(4\pi\epsilon_0 y)((L)/(√(4y^2+L^2))`

`(1)/(4\pi\epsilon_0)=9* 10^9 Nm^2/C^2`

Using the formula

`E=(2* 5.5* 10^(-9)* 9* 10^9* 0.052)/(0.04* √(4(0.04)^2+(0.052)^2))`

`E=1348.8N/C`

c.y=12 cm=
`(12)/(100)=0.12m`

Using the formula

`E=(2* 5.5* 10^(-9)* 9* 10^9* 0.052)/(0.12* √(4(0.12)^2+(0.052)^2))`

`E=174.7N/C`