Question

Calculate the potential difference across the plates of capacitors X and Y.

Hint:

In a series capacitor combination: \[ Q_1=Q_2=Q_3=\cdots \] and \[ V=\frac{Q}{C}. \] Hence the capacitor having smaller capacitance gets a larger share of the voltage.

Answer 1

Concept: When capacitors are connected in series:
• The charge on each capacitor is the same.
• Potential differences divide inversely proportional to capacitances.
• For each capacitor, \[ V=\frac{Q}{C}. \]

Step 1: Calculate the charge stored in the series combination. From part (a), \[ C_{\text{eq}}=4\,\mu\text{F}. \] Battery voltage, \[ V=6\,\text{V}. \] Using \[ Q=C_{\text{eq}}V, \] \[ Q=(4\times10^{-6})(6). \] \[ Q=24\times10^{-6}\,\text{C}. \] \[ \boxed{Q=24\,\mu\text{C}}. \]

Step 2: Calculate voltage across capacitor X. Using \[ V_X=\frac{Q}{C_X}, \] \[ V_X=\frac{24\,\mu\text{C}}{5\,\mu\text{F}}. \] \[ V_X=4.8\,\text{V}. \] Hence, \[ \boxed{V_X=4.8\,\text{V}}. \]

Step 3: Calculate voltage across capacitor Y. Using \[ V_Y=\frac{Q}{C_Y}, \] \[ V_Y=\frac{24\,\mu\text{C}}{20\,\mu\text{F}}. \] \[ V_Y=1.2\,\text{V}. \] Therefore, \[ \boxed{V_Y=1.2\,\text{V}}. \]

Step 4: Verification. The total voltage should equal the battery voltage. \[ V_X+V_Y = 4.8+1.2 = 6.0\,\text{V}. \] This agrees with the applied voltage. Final Answer: \[ \boxed{V_X=4.8\,\text{V}} \] \[ \boxed{V_Y=1.2\,\text{V}} \]