Question

Two like charged, infinitely long parallel wires with the same linear charge density of 3*10-8 c/cm are 2 cm apart. The work done against electrostatic force per unit length to be done in bringing them closer by 1 cm is x/100 j/m: find the value of x

Answer 1

Let $\lambda_1$ and $\lambda_2$ are linear charge densities of two parallel wires.

electric field due to $\lambda_1$ at r distance is given by $E=\frac{\lambda_1}{2\pi\epsilon_0r}$

now, force experienced by wire of linear charge density $\lambda_2$ is $F= qE$

where q = $\lambda_2l$

now, F = $\frac{\lambda_1\lambda_2l}{2\pi\epsilon_0r}$

and force per unit length, $F'=\frac{F}{l}=\frac{\lambda_1\lambda_2}{2\pi\epsilon_0r}$

now , putting $\lambda_1=\lambda_2$ = 3 × 10^-8 C/cm = 3 × 10^-6 C/m , r = 2cm = 0.02m

so, F' = (3 × 10^-6)²/(2 × 3.14 × 8.85 × 10^-12 × 0.02)

= 9 × 10^-12/(6.28 × 8.85 × 2 × 10^-14)

= 0.08 × 10²

= 8 N/m

so, workdone against electrostatic force per unit length = F' × s

= 8N/m × 1cm

= 8 N/m × 0.01m

= 0.08 J/m

= 8/100 J/m

hence, value of x = 8