# Homework Help: Guass's Law

1. Feb 13, 2010

### 05holtel

1. The problem statement, all variables and given/known data

Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.

A small positive point charge of 8 mC moves from a distance of 9 cm to a distance of 17 cm.

How much work is done by the electric field?

Hint: The electric field for a long charged line is:

The Equation is E line = 1/(4piE0)(2lamda/r)

Express the result in the unit mJ and to three significant figures.
2. Relevant equations

The Equation is E line = 1/(4piE0)(2lamda/r)

3. The attempt at a solution

Is this hint equation provided equivalent to Gauss' Law: (2*pi*R*L*E = L*q/eo)

where eo is the coulomb's law constant "epsilon-zero" = 8.89^10^-12 N*m^2/C^2
and q is the charge per unit length

Therefore,

E(R) = q/(2 pi eo R)
Multiply that by Q = 8 mC and integrate with dR from R = 0.09 to 0.17 m

Also I am confused how to integrate this

2. Feb 13, 2010

### xcvxcvvc

$$E = \frac{\lambda}{2 \pi \epsilon_o r}$$
$$W = \int\limits_{R_1}^{R_2} F_e\, dr$$
$$W = \int\limits_{R_1}^{R_2} qE\, dr$$
$$W = \int\limits_{R_1}^{R_2}\frac{q\lambda}{2 \pi \epsilon_o r}\, dr$$
Everything is constant except for the r.
$$W = \frac{q\lambda}{2 \pi \epsilon_o }\int\limits_{R_1}^{R_2}\frac{1}{r}\, dr$$
$$W = \frac{q\lambda}{2 \pi \epsilon_o} ln(r)|\limits_{R_1}^{R_2}$$

3. Feb 14, 2010

### 05holtel

Thanks so much,
Do you know what the wavelength is

4. Feb 14, 2010

### xcvxcvvc

$$\lambda$$ is linear charge density