Question

Find the potential difference $V_a - V_b$ between the points a and b shows in each parts of the figure.

• A. zero, $-\frac{7}{72}V=-10.3 V$
• B. one, $-\frac{72}{7}V=-10.3 V$
• C. one, $-\frac{7}{72}V=-10.3 V$
• D. zero, $-\frac{72}{7}V=-10.3 V$

Answer 1

Answer: (D)

(4)

Answer 2

Part (a):

The circuit in part (a) is interpreted as a balanced Wheatstone bridge of capacitors. Assuming the labels "2V" are typos for "2$\mu$F", all four arms of the bridge (top-left, top-right, bottom-left, bottom-right) have capacitors of 2 $\mu$F.

For a capacitor bridge, the balance condition is $C_{top-left}/C_{bottom-left} = C_{top-right}/C_{bottom-right}$.

Here, $2\mu F / 2\mu F = 2\mu F / 2\mu F$, which simplifies to $1 = 1$. Since the bridge is balanced, the potential difference between points 'a' and 'b' is zero. $V_a - V_b = 0$.

Part (b):

The circuit in part (b) is a combination of capacitors and voltage sources. We can use node analysis to find the potential difference $V_a - V_b$. Let's assume the potential at point 'b' is $0V$ ($V_b = 0$).

The charge on each capacitor is given by $Q = C \Delta V$, where $\Delta V$ is the potential difference across the capacitor. For a capacitor in series with a battery, the potential difference across the capacitor is $V_a - (V_b - E)$ if the positive terminal of the battery is closer to 'b'.

The sum of charges on the isolated node 'a' must be zero in the steady state.

1. Top branch (4 $\mu$F, 6V): The potential at the end of the 6V battery closer to 'b' is $V_b - 6 = 0 - 6 = -6V$. The charge on the 4 $\mu$F capacitor is $Q_1 = 4 \mu F \times (V_a - (-6V)) = 4(V_a + 6)$.
2. Middle branch (2 $\mu$F, 12V): The potential at the end of the 12V battery closer to 'b' is $V_b - 12 = 0 - 12 = -12V$. The charge on the 2 $\mu$F capacitor is $Q_2 = 2 \mu F \times (V_a - (-12V)) = 2(V_a + 12)$.
3. Bottom branch (1 $\mu$F, 24V): The potential at the end of the 24V battery closer to 'b' is $V_b - 24 = 0 - 24 = -24V$. The charge on the 1 $\mu$F capacitor is $Q_3 = 1 \mu F \times (V_a - (-24V)) = 1(V_a + 24)$.

Applying charge conservation at node 'a' (sum of charges on plates connected to 'a' is zero):

$Q_1 + Q_2 + Q_3 = 0$

$4(V_a + 6) + 2(V_a + 12) + 1(V_a + 24) = 0$

$4V_a + 24 + 2V_a + 24 + V_a + 24 = 0$

$7V_a + 72 = 0$

$7V_a = -72$

$V_a = -\frac{72}{7} V$

Since $V_b = 0V$, the potential difference $V_a - V_b = -\frac{72}{7} V$.

Numerically, $-\frac{72}{7} V \approx -10.2857 V \approx -10.3 V$.

Combining the results for (a) and (b), the potential differences are $0$ and $-\frac{72}{7} V$.