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(A) The electric field $\vec{E}$ always points in the direction of decreasing electric potential. Since the potential decreases from $80 	ext{ V}$ to

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Rightward; $4 > 3 > 2 > 1$

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(A) The electric field $\vec{E}$ always points in the direction of decreasing electric potential. Since the potential decreases from $80 	ext{ V}$ to $40 	ext{ V}$ as we move from left to right,the electric field is directed towards the right.The work done in moving a charge $q$ between two points is given by $W = q \Delta V$. For an electron,$q = -e$,so the magnitude of work done is $|W| = e |\Delta V|$.Calculating the potential difference $|\Delta V|$ for each path:Path $1$: Starts at $80 	ext{ V}$,ends at $40 	ext{ V}$. $|\Delta V_1| = |40 - 80| = 40 	ext{ V}$.Path $2$: Starts at $70 	ext{ V}$,ends at $60 	ext{ V}$. $|\Delta V_2| = |60 - 70| = 10 	ext{ V}$.Path $3$: Starts at $80 	ext{ V}$,ends at $60 	ext{ V}$. $|\Delta V_3| = |60 - 80| = 20 	ext{ V}$.Path $4$: Starts at $70 	ext{ V}$,ends at $50 	ext{ V}$. $|\Delta V_4| = |50 - 70| = 20 	ext{ V}$.Comparing the magnitudes: $|\Delta V_1| = 40 	ext{ V}$,$|\Delta V_3| = 20 	ext{ V}$,$|\Delta V_4| = 20 	ext{ V}$,$|\Delta V_2| = 10 	ext{ V}$.Therefore,the order of magnitude of work done is $1 > 3 = 4 > 2$.

The figure shows a family of parallel equipotential surfaces and four paths along which an electron is made to move from one surface to another.
$(I)$ What is the direction of the electric field?
$(II)$ Rank the paths according to the magnitude of work done,greatest first.

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The figure shows a family of parallel equipotential surfaces and four paths along which an electron is made to move from one surface to another. $(I)$ What is the direction of the electric field?$(II)$ Rank the paths according to the magnitude of work done,greatest first.