= Solution
For the <image force on a line charge above a conducting plane>, with no imposed external <electric field>, use the <method of images>: replace the plane by a parallel image line of charge $-\lambda$ at height $-h$. The image is $2h$ from the real line, and its field at the real line has magnitude $\lambda/(4\pi\epsilon_0h)$. Therefore
$$
\boxed{\frac{\mathbf F}{\text{length}}=-\frac{\lambda^2}{4\pi\epsilon_0h}\,\hat{\mathbf z}.}
$$
Here $\hat{\mathbf z}$ points upward from the plane. \b[The force is attractive and normal to the conductor], for either sign of $\lambda$. The image reproduces the induced surface charge field; it is not an extra physical charge, and no factor $1/2$ is inserted in the force on the actual line.
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