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(B) Given: Capacitance of $A$ $(C_A)$ = $2 \, \mu F$,Capacitance of $B$ $(C_B)$ = $3 \, \mu F$,and total voltage $(V)$ = $10 \, V$.Since the capacitor

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The potential difference across $A$ is $6 \, V$ and across $B$ is $4 \, V$.

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(B) Given: Capacitance of $A$ $(C_A)$ = $2 \, \mu F$,Capacitance of $B$ $(C_B)$ = $3 \, \mu F$,and total voltage $(V)$ = $10 \, V$.Since the capacitors are in series,the charge $(Q)$ on both capacitors is the same.For capacitors in series,the potential difference is inversely proportional to the capacitance $(V \propto 1/C)$.Therefore,$V_A / V_B = C_B / C_A = 3 / 2$.Let $V_A = 3x$ and $V_B = 2x$.Since $V_A + V_B = 10 \, V$,we have $3x + 2x = 10 \, V$,which gives $5x = 10 \, V$,so $x = 2 \, V$.Thus,$V_A = 3(2) = 6 \, V$ and $V_B = 2(2) = 4 \, V$.Therefore,the potential difference across $A$ is $6 \, V$ and across $B$ is $4 \, V$.

Two capacitors $A$ and $B$ are connected in series as shown in the figure. When the switch $S$ is closed and both capacitors are fully charged,then:

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