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(D) Given $V_C = 0\,V$ and the battery voltage is $20\,V$ connected between $A$ and $C$ with positive terminal at $A$,so $V_A = 20\,V$.Applying Kirchh

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all of the above.

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(D) Given $V_C = 0\,V$ and the battery voltage is $20\,V$ connected between $A$ and $C$ with positive terminal at $A$,so $V_A = 20\,V$.Applying Kirchhoff's Current Law $(KCL)$ at node $B$:The sum of charges on capacitors connected to node $B$ must be zero (assuming no external charge source at $B$):$q_{AB} + q_{DB} + q_{CB} = 0$$4(V_A - V_B) + 2(V_D - V_B) + 2(V_C - V_B) = 0$Since $V_C = 0$,this becomes:$4(V_A - V_B) + 2(V_D - V_B) - 2V_B = 0$$4(V_A - V_B) + 2(V_D - V_B) = 2V_B$. This matches option $B$.Applying $KCL$ at node $D$:$q_{AD} + q_{BD} + q_{CD} = 0$$2(V_A - V_D) + 2(V_B - V_D) + 4(V_C - V_D) = 0$Since $V_C = 0$,this becomes:$2(V_A - V_D) + 2(V_B - V_D) - 4V_D = 0$$2(V_A - V_D) + 2(V_B - V_D) = 4V_D$. This matches option $C$.Since options $A$,$B$,and $C$ are correct,the correct answer is $D$.

The figure shows a diagonal symmetric arrangement of capacitors and a battery. If the potential of $C$ is zero,then

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