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Solution :

When the capacitors are connected in parallel (Fig. 24.9), all their interconnected plates have the same potential, therefore `q_(1)=C_(1)( varphi_(1)-varphi_(2)) and q_(2)=C_(2)(varphi_(1)-varphi_(2))`, where `q_(1) and q_(2)` are the charges of the respective capacitors. The charge of the system is <br> `q=q_(1)+q_(2)=(C_(1)+C_(2)) (varphi_(1)-varphi_(2))`. <br> On the other h and `q=C(varphi_(1)-varphi_(2))`, where C is the equivalent capacitance. Hence the formula sought. <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/ARG_AAP_PIP_PHY_C24_E01_010_S01.png" width="80%">