Question

A 2-m long thin and straight metal wire carries a positive charge of 3 µ C. (a) What is the linear charge density on the wire? (b) Hence, find the electric intensity at a point 1.5 m away from the centre of the wire.

Answer 1

Given:

• Length L = 2 m
• Charge Q = 3µC = 3 x 10−6 C
• Point is at r = 1.5 m from the centre
• k = `1/(4pi∈_0) ≈ 9 xx 10^9 (N.m^2)/C^2`

Linear charge density

`lambda = Q/L = (3 xx 10^-6)/(2) = 1.5 xx 10^-6` C/M

λ = 1.5 µC/m

Electric field at r = 1.5 m (point on perpendicular bisector)

For a finite straight wire of length L:

E = `1/(4pi∈_0) * (2lamdaL)/(rsqrt(r^2 + (L/2)^2))`

Plug in numbers:

2λL = 2(1.5 × 10⁻⁶)(2) = 6 × 10⁻⁶

`rsqrt(r^2 + (L/2)^2) = 1.5sqrt(1.5^2 + 1^2) = 1.5sqrt(3.25) ≈ 2.704`

`E = 9 xx 10^9 * (6 xx 10^-6)/(2.704) = (5.4 xx 10^4)/(2.704) ≈ 2.0 xx 10^4` N/C

E ≈ 2.0 × 10⁴ N/C