A parallel plate air capacitor is connected to a battery. If plates of the capacitor are pulled further apart, then which of the following statement(s) is/are correct?

When the battery is disconnected, then the charge will remain constant. Now when the plates are pulled apart, then we have

$E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$

So $E$ remains constant and hence option $\left(1\right)$ is correct. Work is done against attractive force by the external force $F_{ext}$

So option $\left(2\right)$ is also correct

Since $U=\frac{1}{2}C{V}^2$ when the battery stays connected to the capacitor then $V=$constant. When the separation between the plates is increased, the capacitance given by $C=\frac{{\epsilon }_{0}A}{d}$ decreases and hence option $\left(3\right)$ is also correct