The force $\displaystyle \vec{F}$ on a given charge $\displaystyle q$ in the presence of an electric field $\displaystyle \vec{E}$ is given below as:

$\displaystyle \vec{F}=q\vec{E}$

It is important to note that the force will be directed along the same direction as the electric field.  Therefore, for our problem, we are able to write

$\displaystyle \vec{F}=\left(3e^+ \right )\left(500\hat{y} \right )\left(\frac{N}{C} \right )$

$\displaystyle \vec F=3\left(1.602 * 10^{-19}C \right )500 \hat{y} \frac{N}{C}=2.40 \times 10^{-16}\hat{y}(N)$

The definition of electric force on a charge in an electric field is:

$\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

$\displaystyle \overrightarrow{F}=< .005,.005,.003>\frac{N}{C}*(6*10^{-6}C)$

$\displaystyle \overrightarrow{F}=< 3*10^{-8},3*10^{-8}, 1.8*10^{-8}>N$

The definition of electric force on a charge in an electric field is:

$\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

$\displaystyle \overrightarrow{F}=< -.008,-.008,.003>\frac{N}{C}*(9*10^{-9}C)$

$\displaystyle \overrightarrow{F}=< -7.2*10^{-11},-7.2*10^{-11}, 2.7*10^{-11}>N$

The definition of electric force on a charge in an electric field is:

$\displaystyle \overrightarrow{F}=\overrightarrow{E}q$

$\displaystyle \overrightarrow{F}=< .007,.006,.019>\frac{N}{C}*(35*10^{-9}C)$

$\displaystyle \overrightarrow{F}=< 2.45*10^{-10},2.1*10^{-10}, 6.65*10^{-10}>N$

Use the following form of the electric force equation:

$\displaystyle F_E=Eq$

Plug in values:

$\displaystyle F_E=-33*10^{-9}*< 2.5,3.1>\frac{N}{C}$

$\displaystyle F_E=< -8.25*10^{-9},-1.02*10^{-7}>N$

Use the following equation for electric force:

$\displaystyle F_E=Eq$

Plug in values:

$\displaystyle 50*10^{-6}*1.1=60*10^{-9}*E$

$\displaystyle E=9.16*10^3\frac{N}{C}$

Using $\displaystyle F_E=Eq$ and $\displaystyle F=ma$

Combining equations:

$\displaystyle Eq=ma$

Solving for $\displaystyle E$

$\displaystyle \frac{m*a}{q}=E$

Converting $\displaystyle \mu g$ to $\displaystyle kg$, $\displaystyle nC$ to $\displaystyle C$ and plugging in values

$\displaystyle \frac{55*10^{-9}*< 50,0>}{95*10^{-9}}=E$

$\displaystyle E=< 28.9,0>\frac{N}{C}$

As one moves closer to a negative charge, the electric field increases in the direction of the negative charge.

Recall that the formula for the magnitude of force $\displaystyle F$ is given by:

$\displaystyle F=|E*q|$

Where $\displaystyle E$ is the electric field strength and $\displaystyle q$ is the charge. 

Since the electric field strength is constant,

$\displaystyle F=(41 \frac{N}{C})*(1.602*10^{-19}C)=6.6*10^{-18}N$

To determine direction, remember that electrons will move towards the positively charged plates since electrons are negatively charged. 

There will be no work done. The point charge at the center is emitting an electric field towards itself. If the electron is moved in a circle around it, it will have moved perpendicular to the force the entire time, and thus no work has been done.