What amount of heat will be generated in the circuit, after the switch is shifted from position 1 to position 2 (in microjoule )? $(\mathrm{C}={\mathrm{C}}_0=2\mathrm{\mu F},\mathrm{E}=\sqrt{6}\mathrm{V})$

By KVL,

$\frac{q}{C}=\frac{q_0}{C_0} ...\left(1\right)$ 
and  $\frac{q+q_0}{C}-E+\frac{q}{C}=0$
 $\Rightarrow q+q_0-CE+q=0$

$2q+q_0-CE=0$                    …(2)
From (1) and (2),
$q=\frac{C^{2}E}{2C+C_0}$   and  $q_0=\frac{C{C}_{0}E}{2C+C_0}$
By symmetry in the figure, when the switch is shifted from 1 to 2, the new charge distribution will be as shown below.

From above analysis, the charge flown through the battery when the switch is shifted $=q_0$
By symmetry in the figure, initial and final electrostatic potential energy of the system is same.
$W_{Bat}=\text{Heat}+U_2-U_1$
But $U_2=U_1$
So, $W_B=\text{Heat}$

$\text{Heat}=q_{0}E=\frac{C{C}_0{E}^2}{2C+C_0}$
Substituting the values,

$\text{Heat}=\frac{\left(2\times {10}^{-6}\right)\left(2\times {10}^{-6}\right)\times 6}{2\times \left(2\times {10}^{-6}\right)+\left(2\times {10}^{-6}\right)}=4\times {10}^{-6}\text{J}$