Use the equations for potential electric energy and electric potential:

\(\displaystyle PE_{electric} = Vq\)

\(\displaystyle V=Ed\) 

Substitute.

\(\displaystyle PE_{electric} = Edq\)

Plug in known values and solve.

\(\displaystyle PE_{electric}= (2.75\cdot 10^{-9}C)(35\cdot 10^{-6}\frac{N}{C})(3.5\cdot 10^{-3}m)\)

\(\displaystyle PE_{electric} = 337\cdot 10^{-18} Nm = 337\cdot10^{-18} J=3.37\cdot 10^{-16}J\)

Electric field strength is the slope of the electric potential:

\(\displaystyle E=\frac{\Delta V}{\Delta x}\) 

Remember to convert into SI units (meters):

\(\displaystyle E=\frac {V_2-V_1}{\Delta x}\)

\(\displaystyle E=\frac{7V-3V}{0.03m}\)

\(\displaystyle E=133 \frac{V}{m} = 133 \frac{N}{C}\)

This problem is solved using the work-kinetic energy theory: \(\displaystyle \Sigma W=\Delta K\). In an electric field, work is equal to charge times change in potential: \(\displaystyle W=q\Delta V\)

Since an electron has a negative charge, decreasing potential increases its kinetic energy, the opposite of what would happen to a proton with its positive charge. Combine these equations:

\(\displaystyle q\Delta V=K_f-K_i\)

\(\displaystyle q\Delta V=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}\) 

Plug in known values and solve

\(\displaystyle 1.6\times10^{-19}\cdot 1.5=\frac{1}{2}\cdot 9.11\cdot 10^{-31}v_f^{2}-\frac{1}{2}\cdot9.11\cdot10^{-31}\cdot(2.4\cdot10^{4})^{2}\)

\(\displaystyle v_f=3.63\cdot 10^{5}\frac{m}{s}\)

In order to solve for electrical potential energy, we'll need to remember the equation for it.

\(\displaystyle U=k\frac{q_{s}q_{t}}{r}\)

In the above expression, \(\displaystyle U\) represents electrical potential energy, \(\displaystyle q_{s}\) and \(\displaystyle q_{t}\) represent different point charges, and \(\displaystyle r\) represents the distance between their centers. In this example, one of these charges will be the source of the external electric field, while the other charge will be the one that is undergoing a transposition from point A to point B.

Furthermore, we can remember the equation for voltage:

\(\displaystyle V=k\frac{q_{s}}{r}\)

With both these equations in mind, we can combine the two:

\(\displaystyle U=Vq_{t}\)

This above expression tells us that the electrical potential energy of a system is directly proportional to the voltage change and to the charge of the test charge that is undergoing the voltage change.

Plug in the values and solve for electrical potential energy:

\(\displaystyle U=(6V)(1 C)=6J\)

For this question, we're presented with a situation in which a positively charged particle is traveling through an electric field along a defined path. We're then asked to determine the amount of energy we need to input in order to make this process occur.

We'll need to use an equation that relates charge, electric field, and energy, which we can derive as follows.

\(\displaystyle W=Fd\)

\(\displaystyle F=Eq\)

\(\displaystyle W=Eqd\)

Next, we need to make sure to use the correct value for distance. Since we know that the electric force is a conservative force, we know that the movement of the particle from A to B is independent of the path taken. In other words, for a conservative force, the only thing that matters is the initial state and the final state. Also, since the electric field is oriented horizontally from right to left, we only care about the movement of the particle along the horizontal direction. Thus, we do not use the \(\displaystyle 12\: cm\) value (distance of path taken), but instead we use the \(\displaystyle 10\: cm\) value (net distance traveled along the electric field). 

Plugging in the values we have, we can obtain our answer:

\(\displaystyle W=(20\: \frac{N}{C})(5\: C)(0.10\: m)\)

\(\displaystyle W=10\:J\)

Converting units:

\(\displaystyle 50nC=50*10^{-9}C\)

\(\displaystyle 60V=60\frac{N*m}{C}\)

Use the equation for electric potential energy:

\(\displaystyle PE_{elec}=V*q\)

\(\displaystyle PE_{elec}=60*50*10^{-9}\)

\(\displaystyle PE_{elec}=3.0*10^{-6}\textup{ J}\)

Conservation of energy:

\(\displaystyle E_i+W=E_f\)

Assuming no external work done

\(\displaystyle EPE_i=KE_f\)

Electric potential energy:

\(\displaystyle EPE=Vq\)

\(\displaystyle Vq=.5mv^2\)

Solving for velocity:

\(\displaystyle v=\sqrt{\frac{Vq}{.5m}}\)

Plugging in values:

\(\displaystyle v=\sqrt{\frac{1*10^{-9}*1.6*10^{-19}}{.5*9.1*10^{-31}}}\)

 \(\displaystyle v=18.8\frac{m}{s}\)

Conservation of energy:

\(\displaystyle E_i+W=E_f\)

Assuming no external work done

\(\displaystyle EPE_i=KE_f\)

Electric potential energy:

\(\displaystyle EPE=Vq\)

\(\displaystyle Vq=.5mv^2\)

Solving for velocity:

\(\displaystyle v=\sqrt{\frac{Vq}{.5m}}\)

Plugging in values:

\(\displaystyle v=\sqrt{\frac{1*10^{-9}*2*1.6*10^{-19}}{.5*6.6*10^{-27}}}\)

 \(\displaystyle v=.31\frac{m}{s}\)

Conservation of energy:

\(\displaystyle E_i+W=E_f\)

Assuming no external work done

\(\displaystyle EPE_i=KE_f\)

Electric potential energy:

\(\displaystyle EPE=Vq\)

\(\displaystyle Vq=.5mv^2\)

Solving for velocity:

\(\displaystyle v=\sqrt{\frac{Vq}{.5m}}\)

Plugging in values:

\(\displaystyle v=\sqrt{\frac{1*10^{-9}*1.6*10^{-19}}{.5*1.67*10^{-27}}}\)

 \(\displaystyle v=.438\frac{m}{s}\)

Force in an electric field:

\(\displaystyle F_{Electric Field}=Eq\)

Using conservation of energy, assuming no external work on system:

\(\displaystyle Energy_i=Energy_f\)

\(\displaystyle KE_i+EPE_i=KE_f+EPE_f\)

Definition of electric potential energy:

\(\displaystyle V*q=EPE\)

Definition of electric potential:

\(\displaystyle V=E*d\)

Combining equations:

\(\displaystyle .5mv^2_i+E*d_i*q=.5mv^2_f+E*q*d_f\)

Assuming final velocity is zero:

\(\displaystyle .5mv^2_i=E*q*(d_f-d_i)\)

\(\displaystyle (d_f-d_i)\) is the distance traveled

Converting \(\displaystyle \frac{km}{hr}\) to \(\displaystyle \frac{m}{s}\)

\(\displaystyle 300\frac{km}{hr}*\frac{1 hr}{3600s}*\frac{1000m}{1 km}=83.3\frac{m}{s}\)

Plugging in values:

\(\displaystyle .5*1625*83.3^2_i=50*1*10^{-9}*(d_f-d_i)\)

Solving for \(\displaystyle (d_f-d_i)\)

\(\displaystyle (d_f-d_i)=1.127*10^{14}m\)