Find the potential difference between points A and B and between points B and C of the figure in steady state.

Hint: Remember the statement given in the question that circuit is in steady state which means i = 0 A find the net capacitance for AB and BC circuit and then use the formula \[({V_A} - {V_B}){C_1} = ({V_B} - {V_C}){C_2}\] to find the potential difference between A and B also B and C.

Complete Step-by-Step solution:
Let first 3uF of AB circuit be x and y and let 1uf of AC be z
Since, according to the question the circuit is in the steady state, therefore, the current in the circuit is 0 and also the circuit will be as given below

So, since the capacitor x and y are in the parallel formation .i.e. therefore the net capacitance is \[{C_{net}} = 3 + 3 = 6uF\]
The formed circuit is

Now let 1uF of BC circuit be n and m.
Since n and m are also in the parallel formation therefore the net capacitance is \[{C_{net}} = 1 + 1 = 2uF\]
Hence, the circuit will be

So now for the potential difference at point A and B
Since ${V_A} = - 100$
By the formula of \[({V_A} - {V_B}){C_1} = ({V_B} - {V_C}){C_2}\]
\[( - 100 - {V_B})6 = ({V_B} - 0)2\]
${V_B} = - 75$
Therefore \[({V_A} - {V_B}) = - 100 - ( - 75)\]
\[({V_A} - {V_B}) = - 25V\]
For B and C
Since we have ${V_B} = - 75$and ${V_C} = 0$
Therefore the potential difference between B and C is
${V_B} - {V_C} = - 75 - 0$
Hence ${V_B} - {V_C} = - 75V$
Hence the potential difference between A and B is -25V and the potential difference between B and C is - 75V.

Note: The term steady state is a state of circuit in which the charge (or current) flowing into any point in the circuit has to equal the charge (or current) flowing out. A system can achieve a steady state when the current at each point in the circuit is constant (does not change with time).