Question

Find the equivalent capacitance across points E and G.

Answer 1

3C/2

Answer 2

To find the equivalent capacitance across points E and G, we can utilize the symmetry of the circuit. Assume all capacitors have capacitance $C$.

1. Symmetry Argument: The circuit is symmetric about the diagonal line passing through E and G. This implies that points F and H are at the same potential. Similarly, points A and B are at the same potential, and points C and D are at the same potential. However, the connection pattern for A, B, C, D is not simple. A more robust symmetry argument is that if we apply a voltage $V$ at E and $0$ at G, then due to symmetry, the potential at F must be equal to the potential at H ($V_F = V_H$).

2. Simplification due to $V_F = V_H$: Since $V_F = V_H$, no current flows through any capacitor directly connecting F and H (there isn't one in this diagram). More importantly, we can merge points F and H into a single node, let's call it $F'$.

   • Capacitors $C_{EF}$ (labeled '1') and $C_{EH}$ are in parallel between E and $F'$. Their equivalent capacitance is $C_{EF'} = C + C = 2C$.
   • Capacitors $C_{FG}$ and $C_{HG}$ are in parallel between $F'$ and G. Their equivalent capacitance is $C_{F'G} = C + C = 2C$.
   • The capacitor labeled '2' is between the midpoint of the capacitor $C_{EH}$ and the midpoint of $C_{FG}$. This implies it's a bridge element. However, with $V_F = V_H$, the potentials of the points on $C_{EH}$ and $C_{FG}$ that are equidistant from E and G respectively will also be equal. Specifically, the points where capacitor '2' connects are symmetric with respect to the EG diagonal. Let these points be $P_1$ (on $C_{EH}$) and $P_2$ (on $C_{FG}$). Due to symmetry, $V_{P_1} = V_{P_2}$. Therefore, no current flows through capacitor '2', and it can be removed.
   • The other innermost capacitor (vertical one) connects points on $C_{EF}$ and $C_{HG}$. Let these points be $Q_1$ (on $C_{EF}$) and $Q_2$ (on $C_{HG}$). Due to symmetry, $V_{Q_1} = V_{Q_2}$. Therefore, no current flows through this capacitor either, and it can be removed.
   • Connections from the outer square:
     • $C_{CF}$ and $C_{DH}$ are in parallel between C and $F'$, and D and $F'$ respectively. Since F and H are merged, $C_{CF}$ and $C_{DH}$ become $C_{CF'}$ and $C_{DF'}$. If C and D are also symmetric, then $V_C = V_D$.
     • $C_{BG}$ connects B to G.
     • $C_{AE}$ connects A to E.

   Let's re-evaluate the symmetry: The overall network looks like a 2D grid. The symmetry argument $V_F = V_H$ holds if the network is symmetric about the line EG.

   • E-F (C) and E-H (C) are symmetric.
   • F-G (C) and H-G (C) are symmetric.
   • The innermost square with capacitors '1' and '2' (and the others) also exhibits symmetry around the EG axis. The capacitor '2' connects symmetric points, so it can be removed. The other capacitor in the inner square also connects symmetric points, so it can be removed.
   • The connections from F and H to the outer layer: $C_{CF}$ and $C_{DH}$. For $V_F = V_H$ to hold, the rest of the circuit must also be symmetric.
   • Points C and D are symmetric about the EG axis. (C is connected to F, A, B; D is connected to H, A, B).
   • Points A and B are symmetric about the EG axis. (A is connected to C, D, E; B is connected to C, D, G).

   Given this symmetry, we can confirm $V_F = V_H$. Thus, we can merge F and H. The capacitors '1' and '2' and the other two innermost capacitors become irrelevant as their connecting points are at the same potential.

3. Redrawing the simplified circuit:

   • Between E and $F'$: $C_{EF}$ and $C_{EH}$ are in parallel $\implies 2C$.
   • Between $F'$ and G: $C_{FG}$ and $C_{HG}$ are in parallel $\implies 2C$.
   • Between C and $F'$: $C_{CF}$ and $C_{DH}$ are in parallel $\implies 2C$. (Since $V_C=V_D$, we can merge C and D into $C'$).
   • Between A and E: $C_{AE}$.
   • Between B and G: $C_{BG}$.
   • Between A and $C'$: $C_{AC}$ and $C_{AD}$ are in parallel $\implies 2C$.
   • Between B and $C'$: $C_{CB}$ and $C_{DB}$ are in parallel $\implies 2C$.
   • Between A and B: There is no direct connection.

   Let's use a simpler approach based on the symmetry of the nested squares. Consider the innermost square first. The capacitor labeled '2' connects two points on the inner square. The diagram is drawn in a way that suggests a fractal structure. If all capacitors are $C$. The innermost square has 4 capacitors. Let's call the vertices $P_1, P_2, P_3, P_4$. The capacitor labeled '2' is one side, say $C_{P_1P_2}$. The capacitor connected vertically to '2' is $C_{P_3P_4}$. The connections from $E, F, G, H$ to these inner nodes are also capacitors.

   A common interpretation for such nested diagrams is that the network is made of identical unit cells. This is a grid of capacitors. Let's assume the problem is a standard type of problem where the structure is a series of nested squares, and we need to find the equivalent capacitance between diagonally opposite points (E and G).

   Let's assume all capacitors have capacitance $C$. The central capacitor '2' connects the midpoints of $EH$ and $FG$. The other central capacitor connects the midpoints of $EF$ and $HG$. Due to symmetry about the EG diagonal, the potential at the midpoint of $EH$ is equal to the potential at the midpoint of $FG$. Therefore, the capacitor '2' is shorted and can be removed. Similarly, the potential at the midpoint of $EF$ is equal to the potential at the midpoint of $HG$. Therefore, the other central capacitor can also be removed.

   Now, we are left with the outer and middle layers. The outer layer A-C, A-D, D-B, C-B. The middle layer E-F, F-G, G-H, H-E. The connecting capacitors A-E, C-F, B-G, D-H.

   Since we are calculating capacitance between E and G, points A, B, C, D are external. The symmetry $V_F = V_H$ is robust. So, merge F and H into a single node $F'$. The circuit simplifies to:

   1. E to $F'$: $C_{EF}$ in parallel with $C_{EH}$ gives $2C$.
   2. $F'$ to G: $C_{FG}$ in parallel with $C_{HG}$ gives $2C$.
   3. A to E: $C_{AE}$.
   4. C to $F'$: $C_{CF}$.
   5. D to $F'$: $C_{DH}$.
   6. B to G: $C_{BG}$.
   7. A to C: $C_{AC}$.
   8. A to D: $C_{AD}$.
   9. D to B: $C_{DB}$.
   10. C to B: $C_{CB}$.

   Now, let's consider the symmetry of the outer layer. The diagonal EG is the axis of symmetry. A and B are symmetric. C and D are symmetric. So, $V_A = V_B$ and $V_C = V_D$. We can merge A and B into $A'$. We can merge C and D into $C'$.

   Simplified connections:

   • E to $F'$: $2C$.
   • $F'$ to G: $2C$.
   • $A'$ to E: $C_{AE}$ and $C_{BG}$ are effectively parallel. So $2C$.
   • $C'$ to $F'$: $C_{CF}$ and $C_{DH}$ are effectively parallel. So $2C$.
   • $A'$ to $C'$: $C_{AC}$ and $C_{AD}$ and $C_{CB}$ and $C_{DB}$ are effectively parallel. No, this is incorrect. If A and B are merged, and C and D are merged, then $C_{AC}$ and $C_{AD}$ are between $A'$ and $C'$. $C_{CB}$ and $C_{DB}$ are between $A'$ and $C'$. So, between $A'$ and $C'$: $C_{AC} + C_{AD} + C_{CB} + C_{DB} = 4C$.

   Let's redraw the simplified circuit with merged nodes: Nodes: E, G, $F'$, $A'$, $C'$. Capacitors:

   1. $C_{EF'} = 2C$ (from E to $F'$)
   2. $C_{F'G} = 2C$ (from $F'$ to G)
   3. $C_{A'E} = 2C$ (from $A'$ to E) - this is $C_{AE}$ and $C_{BG}$ combined.
   4. $C_{C'F'} = 2C$ (from $C'$ to $F'$) - this is $C_{CF}$ and $C_{DH}$ combined.
   5. $C_{A'C'} = 4C$ (from $A'$ to $C'$) - this is $C_{AC}$, $C_{AD}$, $C_{CB}$, $C_{DB}$ combined.

   Now, we have a network of 5 nodes. We need $C_{EG}$. The path E-$F'$-G is $2C$ in series with $2C$, which gives an equivalent of $C$. This path is in parallel with other paths.

   Let's use the symmetry argument more carefully. The line EG is an axis of symmetry. $V_F = V_H$. $V_A = V_B$. $V_C = V_D$. Also, the innermost capacitors are shorted because their connecting points are symmetrical.

   Consider the circuit as a combination of sections.

   1. Inner square EFGH: With F and H shorted, and the innermost capacitors removed.

      • $C_{EF}$ and $C_{EH}$ are in parallel ($2C$) between E and F/H.
      • $C_{FG}$ and $C_{HG}$ are in parallel ($2C$) between F/H and G.
      • These two $2C$ capacitors are in series. $C_{E \to G \text{ via F/H}} = (2C \times 2C) / (2C + 2C) = C$.

   2. Outer connections:

      • The capacitor $C_{AE}$ connects E to A.
      • The capacitor $C_{BG}$ connects G to B.
      • The capacitor $C_{CF}$ connects C to F.
      • The capacitor $C_{DH}$ connects D to H.

   Since $V_A = V_B$, we can connect A and B by a wire. Since $V_C = V_D$, we can connect C and D by a wire. Since $V_F = V_H$, we can connect F and H by a wire.

   Let's redraw the circuit with merged nodes: E and G are terminals. Node $F'$ (F and H merged). Node $A'$ (A and B merged). Node $C'$ (C and D merged).

   Connections:

   • E to $F'$: $C_{EF}$, $C_{EH}$ $\implies 2C$.
   • $F'$ to G: $C_{FG}$, $C_{HG}$ $\implies 2C$.
   • E to $A'$: $C_{AE}$.
   • G to $A'$: $C_{BG}$. (These two are in parallel between E/G and $A'$).
   • $F'$ to $C'$: $C_{CF}$, $C_{DH}$ $\implies 2C$.
   • $A'$ to $C'$: $C_{AC}$, $C_{AD}$, $C_{CB}$, $C_{DB}$. $C_{AC}$ is between A and C. $C_{AD}$ is between A and D. $C_{CB}$ is between C and B. $C_{DB}$ is between D and B. So, $C_{AC}$ (between $A'$ and $C'$), $C_{AD}$ (between $A'$ and $C'$), $C_{CB}$ (between $C'$ and $A'$), $C_{DB}$ (between $C'$ and $A'$). All these 4 capacitors are in parallel between $A'$ and $C'$. So $4C$.

   The equivalent circuit becomes:

   Where: $C_{EF'} = 2C$ $C_{F'G} = 2C$ $C_{AE\_BG} = C_{AE} + C_{BG} = 2C$ (between E and A', and G and A') $C_{CF\_DH} = C_{CF} + C_{DH} = 2C$ (between C' and F') $C_{AC\_AD\_CB\_DB} = C_{AC} + C_{AD} + C_{CB} + C_{DB} = 4C$

   This is a complex circuit. Let's try to simplify it using potentials. Let $V_E = V$, $V_G = 0$. Let $V_{F'} = x$, $V_{A'} = y$, $V_{C'} = z$.

   Current conservation at $F'$: $(V - x) 2C + (0 - x) 2C + (z - x) 2C = 0$ $2V - 2x - 2x + 2z - 2x = 0$ $2V + 2z - 6x = 0 \implies V + z = 3x$ (Eq. 1)

   Current conservation at $A'$: $(V - y) C_{AE} + (0 - y) C_{BG} + (z - y) 4C = 0$ $(V - y) C + (0 - y) C + (z - y) 4C = 0$ (Assuming $C_{AE}=C_{BG}=C$) $V - y - y + 4z - 4y = 0$ $V + 4z = 6y$ (Eq. 2)

   Current conservation at $C'$: $(x - z) 2C + (y - z) 4C = 0$ $2x - 2z + 4y - 4z = 0$ $2x + 4y - 6z = 0 \implies x + 2y = 3z$ (Eq. 3)

   We have a system of 3 linear equations:

   1. $3x - z = V$
   2. $6y - 4z = V$
   3. $x + 2y - 3z = 0$

   From (3), $x = 3z - 2y$. Substitute $x$ into (1): $3(3z - 2y) - z = V$ $9z - 6y - z = V \implies 8z - 6y = V$ (Eq. 4)

   Now we have (2) and (4) for $y$ and $z$: $6y - 4z = V$ $-6y + 8z = V$ Add them: $4z = 2V \implies z = V/2$.

   Substitute $z = V/2$ into (4): $8(V/2) - 6y = V$ $4V - 6y = V$ $3V = 6y \implies y = V/2$.

   Substitute $y = V/2$ and $z = V/2$ into (3): $x + 2(V/2) - 3(V/2) = 0$ $x + V - 3V/2 = 0$ $x - V/2 = 0 \implies x = V/2$.

   So, $V_{F'} = V/2$, $V_{A'} = V/2$, $V_{C'} = V/2$. This means all the merged central nodes are at the same potential $V/2$. This implies that the capacitors connecting these nodes ($C_{A'C'}$ and $C_{C'F'}$) are shorted (no current flows through them). This simplifies the circuit even further.

   If $V_{A'} = V_{C'} = V_{F'} = V/2$, then:

   • Current from E: $I_E = (V_E - V_{F'}) C_{EF'} + (V_E - V_{A'}) C_{AE\_BG}$ $I_E = (V - V/2) 2C + (V - V/2) 2C$ $I_E = (V/2) 2C + (V/2) 2C = VC + VC = 2VC$.
   • Current into G: $I_G = (V_{F'} - V_G) C_{F'G} + (V_{A'} - V_G) C_{AE\_BG}$ $I_G = (V/2 - 0) 2C + (V/2 - 0) 2C$ $I_G = (V/2) 2C + (V/2) 2C = VC + VC = 2VC$.

   The total current flowing from E (and into G) is $2VC$. Equivalent capacitance $C_{eq} = Q/V = I \times t / V$. $C_{eq} = I_{total} / (dV/dt)$. If we apply DC voltage, $I = C_{eq} \times (dV/dt)$. For equivalent capacitance, $C_{eq} = Q/V_{applied}$. $Q = I \times t$ (if I is constant current). Or $Q = C_{eq} V$.

   Let's consider the total charge entering E is $Q_E$. $Q_E = (V_E - V_{F'}) 2C + (V_E - V_{A'}) 2C$ $Q_E = (V - V/2) 2C + (V - V/2) 2C = (V/2) 2C + (V/2) 2C = VC + VC = 2VC$. The equivalent capacitance $C_{EG} = Q_E / V = 2VC / V = 2C$.

   Let's double-check this result. If $V_{A'} = V_{C'} = V_{F'} = V/2$, it means these three nodes are at the same potential. This implies that the capacitors connecting these nodes ($C_{A'C'}$ and $C_{C'F'}$) have no potential difference across them, so they carry no charge and can be removed from the circuit. The simplified circuit then becomes:

   • E to $F'$: $2C$
   • $F'$ to G: $2C$
   • E to $A'$: $2C$ (This is $C_{AE}$ and $C_{BG}$ combined, but $C_{BG}$ connects to G, not E. This is where the merging of A and B is problematic in this direct interpretation.)

   Let's re-evaluate the symmetry: The plane passing through E and G is a plane of symmetry. This implies $V_F = V_H$. Also, the plane perpendicular to EG and passing through the center of the figure is a plane of symmetry. This implies $V_A = V_B$ and $V_C = V_D$. And the innermost capacitors are shorted.

   So, we have:

   1. Between E and F/H: $2C$ (from $C_{EF}$ and $C_{EH}$)
   2. Between F/H and G: $2C$ (from $C_{FG}$ and $C_{HG}$)
   3. Between E and A/B: $C_{AE}$
   4. Between G and A/B: $C_{BG}$
   5. Between F/H and C/D: $C_{CF}$ and $C_{DH}$ are in parallel, so $2C$.
   6. Between A/B and C/D: $C_{AC}$, $C_{AD}$, $C_{CB}$, $C_{DB}$. $C_{AC}$ is between A and C. $C_{AD}$ is between A and D. $C_{CB}$ is between C and B. $C_{DB}$ is between D and B. So, $C_{AC}$ and $C_{AD}$ are between A and C/D. $C_{CB}$ and $C_{DB}$ are between B and C/D. Since A and B are merged, and C and D are merged, these 4 capacitors are between A/B and C/D, so $4C$.

   Let $V_E = V$, $V_G = 0$. Let $V_{F/H} = x$. Let $V_{A/B} = y$. Let $V_{C/D} = z$.

   Equations from current conservation: At F/H: $(V - x)2C + (0 - x)2C + (z - x)2C = 0 \implies V + z = 3x$ (Eq. 1) At A/B: $(V - y)C + (0 - y)C + (z - y)4C = 0 \implies V + 4z = 6y$ (Eq. 2) At C/D: $(x - z)2C + (y - z)4C = 0 \implies x + 2y = 3z$ (Eq. 3)

   These are the same equations as before, leading to $x = V/2$, $y = V/2$, $z = V/2$. This means all the intermediate nodes ($F/H$, $A/B$, $C/D$) are at the same potential $V/2$. When multiple nodes are at the same potential, any capacitors connecting them are effectively removed (no current flows). So, $C_{A/B, C/D}$ ($4C$) and $C_{C/D, F/H}$ ($2C$) are removed.

   The simplified circuit then consists only of paths from E and G to the common potential $V/2$. From E to $V/2$:

   • $C_{EFH} = 2C$
   • $C_{AE} = C$ (This connects E to A/B, which is at $V/2$) Total capacitance from E to $V/2$ is $2C + C = 3C$.

   From G to $V/2$:

   • $C_{FGH} = 2C$
   • $C_{BG} = C$ (This connects G to A/B, which is at $V/2$) Total capacitance from G to $V/2$ is $2C + C = 3C$.

   The total equivalent capacitance $C_{EG}$ is the series combination of these two effective capacitances. $C_{EG} = (3C \times 3C) / (3C + 3C) = 9C^2 / 6C = 3C/2$.

   Let's re-check the current calculation with the potentials $V_E = V, V_G = 0, V_{F/H} = V/2, V_{A/B} = V/2, V_{C/D} = V/2$. Current from E: $I_E = I_{E \to F/H} + I_{E \to A/B}$ $I_E = (V_E - V_{F/H}) 2C + (V_E - V_{A/B}) C$ $I_E = (V - V/2) 2C + (V - V/2) C = (V/2) 2C + (V/2) C = VC + VC/2 = 3VC/2$.

   The total charge supplied by the source connected across E and G is $Q = C_{eq} V$. Also, $Q = \sum Q_i$. The charge on the equivalent capacitor is $Q_E = 3VC/2$. So $C_{eq} = Q_E / V = (3VC/2) / V = 3C/2$.

   This result seems consistent.

Final check on the symmetry argument: The circuit has mirror symmetry about the plane containing EG and perpendicular to the page. This implies $V_F = V_H$. The circuit also has mirror symmetry about the plane bisecting the angle FEH and perpendicular to the page. This implies $V_A = V_B$ and $V_C = V_D$. Due to this high degree of symmetry, we found that all intermediate nodes (F/H, A/B, C/D) are at the same potential $V/2$ (midway between E and G). This means that any capacitor connecting these intermediate nodes effectively carries no current and can be removed. The capacitors removed are:

• The innermost horizontal capacitor ('2').
• The innermost vertical capacitor.
• The four capacitors $C_{AC}$, $C_{AD}$, $C_{CB}$, $C_{DB}$ (between A/B and C/D).
• The two capacitors $C_{CF}$, $C_{DH}$ (between C/D and F/H).

The remaining capacitors are:

• $C_{EF}$ and $C_{EH}$ (in parallel, $2C$) from E to F/H.
• $C_{FG}$ and $C_{HG}$ (in parallel, $2C$) from F/H to G.
• $C_{AE}$ (from E to A/B).
• $C_{BG}$ (from G to A/B).

The simplified circuit: E is connected to F/H (at $V/2$) by $2C$. E is connected to A/B (at $V/2$) by $C$. So, from E to the common potential $V/2$, we have $2C$ and $C$ in parallel, giving $3C$.

G is connected to F/H (at $V/2$) by $2C$. G is connected to A/B (at $V/2$) by $C$. So, from G to the common potential $V/2$, we have $2C$ and $C$ in parallel, giving $3C$.

Now, the total circuit is equivalent to two capacitors, each of $3C$, connected in series between E and G, with their common point at $V/2$. The equivalent capacitance of two $3C$ capacitors in series is: $C_{eq} = \frac{(3C)(3C)}{3C + 3C} = \frac{9C^2}{6C} = \frac{3C}{2}$.

The problem states "all capacitors" without specifying their values. Usually, in such problems, it implies all capacitors are identical. If they are not identical, the problem cannot be solved without their values. Assuming all capacitors have capacitance $C$.

The final answer is $\frac{3C}{2}$.

The problem does not specify the value of C. So the answer should be in terms of C.

The problem is part (c) of a larger problem. Usually, in such problems, the capacitance is given in the problem statement or implied to be 'C'. Assuming standard capacitance C.

Final Answer: The equivalent capacitance across points E and G is $\frac{3C}{2}$.

The question asks for the equivalent capacitance. There are no options given, so it's a descriptive answer.

The value of C is not given. So the answer is in terms of C.

Let's assume the question is from a context where C is the capacitance of each individual capacitor.

Final Answer seems to be $3C/2$.