Figure shows two identical capacitors, \( C_{1} \) and \( \mathrm{C}_{2} \) each of \( 1 \mu \ma...

Figure shows two identical capacitors, \( C_{1} \) and \( \mathrm{C}_{2} \) each of \( 1 \mu \mathrm{F} \) capacitance connected to a battery of \( 6 \mathrm{~V} \). Initially switch ' \( S^{\prime} \) is closed. After sometime ' \( S \) ' is left open and dielectric slabs of dielectric constant \( k=3 \) are inserted to fill completely the space between the plates of the two capacitors.

How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted.

[ranbirrandhawa]

Sir on inserting dielectric slab charge on c1 becomes 18 microcoulomb, then how come the charge on the second capacitor didn't charge?