How much work would be done by the electric field if $Q = 20C$ were moved from $A\left( {8V} \right)$ to $B\left( {3.5V} \right)$?

Hint: First we need to find the difference in the potential energy by subtracting the initial potential energy from the final potential energy. Then we have to use the formula of work done by the electric field on the given charge which shows the relation between the charge work done and potential energy. Where we know two of the parameters value and from that we can calculate the work done.

Complete step by step answer:
As per the given problem we have $Q = 20C$ which moved from $A\left( {8V} \right)$ to $B\left( {3.5V} \right)$.
We need to calculate the work done by the electric field.
We have,
Change, $Q = 20C$
Final potential energy, ${V_f} = 3.5V$
Initial potential energy, ${V_i} = 8V$
Hence the difference in potential energy will be final potential minus initial potential.
Mathematically,
$\Delta V = {V_f} - {V_i}$
Now putting the respective values in the above equation we will get,
$\Delta V = 3.5V - 8V = - 4.5V$
Now using work done by electric field formula we will get,
$\Delta W = Q\Delta V$
Now putting the known values in the above formula respectively we will get,
$\Delta W = 20C \times \left( { - 4.5V} \right) = - 90J$
Hence the work done by the electric field when the charge moves from $A\left( {8V} \right)$ to $B\left( {3.5V} \right)$ we will get $ - 90J$.
Joule is the SI unit of work done.

Note: Remember if the speed of change is slow down then the electric field is doing negative work. And when the work is done by electric field there is a decrease in potential energy of the moving charge from one point to another and in this given problem our potential energy decreases due to which shows negative work done.