Answer
(a) a potential function for ${\bf{F}}$ is $ - 200\ln \left( {{x^2} + {z^2}} \right)$.
(b) the electron's speed at $Q$ is ${v_Q} \approx 1.59 \times {10^7}$ m/s.
Work Step by Step
(a) We have an electric field (in newtons per coulomb) given by
${\bf{F}}\left( {x,y,z} \right) = 400{\left( {{x^2} + {z^2}} \right)^{ - 1}}\left( {x,0,z} \right)$
Since the electric field is conservative, let the potential energy for the electric force ${q_e}{\bf{F}}$ be $V\left( {x,y,z} \right)$ such that ${q_e}{\bf{F}} = - \nabla V = - \left( {\dfrac{{\partial V}}{{\partial x}},\dfrac{{\partial V}}{{\partial y}},\dfrac{{\partial V}}{{\partial z}}} \right)$. So,
$\nabla V = - 400{q_e}{\left( {{x^2} + {z^2}} \right)^{ - 1}}\left( {x,0,z} \right)$
1. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives
$V\left( {x,y,z} \right) = - 400{q_e}\smallint \dfrac{x}{{{x^2} + {z^2}}}{\rm{d}}x = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + f\left( {y,z} \right)$
2. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives
$V\left( {x,y,z} \right) = - 400{q_e}\smallint 0{\rm{d}}y = 0 + g\left( {x,z} \right)$
3. taking the integral of $\dfrac{{\partial V}}{{\partial z}}$ with respect to $z$ gives
$V\left( {x,y,z} \right) = - 400{q_e}\smallint \dfrac{z}{{{x^2} + {z^2}}}{\rm{d}}x = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + h\left( {x,y} \right)$
Since the three ways of expressing $V\left( {x,y,z} \right)$ must be equal, we get
$ - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + f\left( {y,z} \right) = g\left( {x,z} \right) = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + h\left( {x,y} \right)$
From here we conclude that the potential energy is $V\left( {x,y,z} \right) = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + C$, where $C$ is a constant. Thus, a potential function for ${\bf{F}}$ is $ - 200\ln \left( {{x^2} + {z^2}} \right)$.
(b) By the Law of Conservation of Energy, the total energy $E$ is constant and therefore $E$ has the same value when the electron is at $P = \left( {5,3,7} \right)$ and at $Q = \left( {1,1,1} \right)$.
So,
$E = \dfrac{1}{2}{m_e}{v_P}^2 + {V_P} = \dfrac{1}{2}{m_e}{v_Q}^2 + {V_Q}$
where ${v_P}$ and ${V_P}$ are the velocity and potential energy of the electron at $P$, respectively. Likewise, ${v_Q}$ and ${V_Q}$ are the velocity and potential energy of the electron at $Q$, respectively.
Since ${v_P} = 0$, we obtain
$\dfrac{1}{2}{m_e}{v_Q}^2 = {V_P} - {V_Q}$
${v_Q} = \sqrt {\dfrac{2}{{{m_e}}}\left( {{V_P} - {V_Q}} \right)} $
${v_Q} = \sqrt {\dfrac{2}{{{m_e}}}\left[ { - 200{q_e}\ln \left( {{x_P}^2 + {z_P}^2} \right) + 200{q_e}\ln \left( {{x_Q}^2 + {z_Q}^2} \right)} \right]} $
$ = \sqrt {400\dfrac{{{q_e}}}{{{m_e}}}\left[ { - \ln \left( {25 + 49} \right) + \ln \left( {1 + 1} \right)} \right]} $
$ = \sqrt {400\left( { - 1.76 \times {{10}^{11}}} \right)\cdot\left[ { - \ln \left( {25 + 49} \right) + \ln \left( {1 + 1} \right)} \right]} $
$ \approx 1.59 \times {10^7}$
So, the electron's speed at $Q$ is ${v_Q} \approx 1.59 \times {10^7}$ m/s.