Answer
a). $q=2.7\times10^{-6}C$
b). The charge is positive and the work done to move the charge is also positive.
$W=\Delta V\times q$
Thus, $\Delta V$ is positive.
which means the charge was moved from lower potential to higher potential, i.e. it was moved from negative plate to positive plate.
Work Step by Step
a). W=$+1.6\times10^{-5}J$
$\Delta V=6V$
work done = $W=\Delta V \times q$
Hence, $q=W/\Delta V=\frac{1.6\times10^{-5}}{6}=2.7\times10^{-6}C$
b). The charge is positive and the work done to move the charge is also positive.
$W=\Delta V\times q$
Thus, $\Delta V$ is positive.
which means the charge was moved from lower potential to higher potential, i.e. it was moved from negative plate to positive plate.
