For the given circuit arrangement, find the change on the capacitor in steady state.

The correct option is 75/8 μF (note: the magnitude of charge on the capacitor is 75/8 μC).

Step 1: Analyze the circuit components from the given diagram.

From the image, we can see four parallel branches connected across common potential points (let us call the left side node A and the right side node B):
1. Top branch: A DC voltage source of $V=2.5\text{V}$ with its longer bar (positive terminal) on the left, connected in series with a resistor of resistance $R_1=1\Omega$.
2. Second branch: A capacitor of capacitance $C=5\mathrm{\mu F}$ connected in series with a resistor of resistance $R_2=2\Omega$.
3. Third branch: An ideal diode pointing to the right (forward biased from left to right) connected in series with a resistor of resistance $R_3=3\Omega$.
4. Fourth branch: An ideal diode pointing to the left (reverse biased for current flowing from left to right) connected in series with a resistor of resistance $R_4=4\Omega$.

Step 2: Understand the state of each branch in DC steady state.

In steady state:
- The capacitor acts as an open circuit (no steady-state current flows through the capacitor branch). Therefore, current in the second branch is $I=0$.
- The battery forces current to flow from the positive terminal (left side) toward the right side.
- The diode in the third branch allows current to flow from left to right, so it is forward biased and acts as a closed switch (short circuit with zero resistance).
- The diode in the fourth branch blocks current from flowing from left to right, so it is reverse biased and acts as an open circuit.

Step 3: Calculate the steady-state current in the circuit.

The active conducting loop consists of the top branch (battery of $2.5\text{V}$ and $1\Omega$ resistor) and the third branch ($3\Omega$ resistor).
The total resistance of this closed conducting loop is:

$R_{\text{total}}=1\Omega +3\Omega =4\Omega$

The steady-state current flowing through the main circuit is given by Ohm's Law:

$I=\frac{V}{R_{\text{total}}}=\frac{2.5}{4}\text{A}=\frac{5}{8}\text{A}$

Step 4: Find the potential difference across the capacitor.

Let the potential of node A (left node) be $V_A$ and node B (right node) be $V_B$.
Since current $I$ flows from node A to node B through the third branch containing the $3\Omega$ resistor, the potential difference across nodes A and B is:

$V_{AB}=V_A-V_B=I\times 3\Omega$

${\mathrm{V\}\{AB\}}}_{}=\left(\frac{5}{8}\right)\times 3=\frac{15}{8}\text{V}$

Since no current flows through the $2\Omega$ resistor connected in series with the capacitor, there is no potential drop across the $2\Omega$ resistor. Thus, the potential difference across the capacitor is equal to ${\mathrm{V\}\{AB\}}}_{}=\frac{15}{8}\text{V}$.

Step 5: Calculate the charge on the capacitor.

The charge stored on the capacitor in steady state is given by:

$Q=C\times {\mathrm{V\}\{AB\}}}_{}$

$Q=5\mathrm{\mu F}\times \frac{15}{8}\text{V}=\frac{75}{8}\mathrm{\mu C}$

Thus, the magnitude of the charge on the capacitor in steady state is 75/8 μC (expressed in option units as 75/8 μF).