Question

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x = 2 L is (500 N/C)ˆı. Find the electric field at x = 3 L. The Coulomb constant is 8.99 × 109 N · m2 /C 2 . Answer in units of N/C.

Answer 1

The electric field at x = 3L due to a uniform line charge can be found using the inverse square law for electric fields of point charges. It is calculated that the electric field at x = 3L is 222.22 N/C, which is less than the field at x = 2L due to the increased distance from the charge distribution.

Step-by-step explanation:

To find the electric field at x = 3L due to a uniform line charge of density λ on the x-axis between x = 0 and x = L with a total charge of 7 nC, we note that the electric field at any point outside the charge distribution is equivalent to that of a point charge at the center of the line charge with the same total charge. Since the electric field at x = 2L is given as 500 N/C, we can infer that at x = 3L, which is further away from the charge distribution, the electric field magnitude will be less because the electric field due to a point charge decreases as the square of the distance from the charge. Using the inverse square law for the electric field of a point charge (E ∝ 1/r2), and knowing the relationship between the electric field values at x = 2L and x = 3L, we can find the new electric field value by multiplying the initial electric field value by the square of the ratio of the distances (2L/3L)2.

Therefore, the electric field at x = 3L is given by:

E3L = E2L * (2L/3L)2
E3L = 500 N/C * (2/3)2
E3L = 500 N/C * 4/9
E3L = 222.22 N/C

So, the electric field at x = 3L is 222.22 N/C.

Answer 2

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is
`\lambda`

Total charge, Q = 7 nC =
`7* 10^(- 9) C`

At x = 2L,

Electric field,
`\vec{E_(2L)} = 500N/C`

Coulomb constant, K =
`8.99* 10^(9) N.m^(2)/C^(2)`

Now, we know that:

`\vec{E} = K(Q)/(x^(2))`

Also the line charge density:

`\lambda = (Q)/(L)`

Thus

Q =
`\lambda L`

Now, for small element:

`d\vec{E} = K(dq)/(x^(2))`

`d\vec{E} = K(\lambda )/(x^(2))dx`

Integrating both the sides from x = L to x = 2L

`\int_(0)^(E)d\vec{E_(2L)} = K\lambda \int_(L)^(2L)(1)/(x^(2))dx`

`\vec{E_(2L)} = K\lambda[(- 1)/(x)]_(L)^(2L)] = K(Q)/(L)[frac{1}{2L}]`

`\vec{E_(2L)} = (9* 10^(9))(7* 10^(- 9))/(L)[frac{1}{2L}] = (63)/(L^(2))`

Similarly,

For the field in between the range 2L< x < 3L:

`\int_(0)^(E)d\vec{E} = K\lambda \int_(2L)^(3L)(1)/(x^(2))dx`

`\vec{E} = K\lambda[(- 1)/(x)]_(2L)^(3L)] = K(Q)/(L)[frac{1}{6L}]`

`\vec{E} = (9* 10^(9))(7* 10^(- 9))/(L)[frac{1}{6L}] = (63)/(6L^(2))`

Now,

If at x = 2L,

`\vec{E_(2L)} = 500 N/C`

Then at x = 3L:

`\frac{\vec{E_(2L)}}{3} = (500)/(3) = 166.67 N/C`