Question

The conductor from point A splits into three parallel branches, which then converge to point B. This configuration indicates that the three branches are connected in parallel between A and B. Therefore, the equivalent capacitance between A and B is the sum of the capacitances of the three individual branches. Due to the symmetry of the diagram, it is assumed that all three branches have the same capacitance, say $C$. Thus, the equivalent capacitance $C_{eq} = C + C + C = 3C$.

The diagram shows that the parallel conductors in each branch are separated by a distance $d$. This suggests that each branch acts as a capacitive element, and its capacitance depends on the separation $d$. For a parallel plate capacitor, the capacitance is given by $C = \frac{\epsilon_0 A}{d}$, where $\epsilon_0$ is the permittivity of free space and $A$ is the area of the plates. Assuming that each parallel branch can be modeled as a capacitor with an effective plate area $A$ and separation $d$, the capacitance of each branch is $C = \frac{\epsilon_0 A}{d}$.

Substituting this expression for $C$ into the equation for the equivalent capacitance, we get: $C_{eq} = 3C = 3 \left(\frac{\epsilon_0 A}{d}\right) = \frac{3\epsilon_0 A}{d}$

Answer 1

The equivalent capacitance between points A and B is $\frac{3\epsilon_0 A}{d}$, where $A$ is the effective area of the parallel conductors in each branch and $d$ is the separation between them.

Answer 2

The conductor from point A splits into three parallel branches, which then converge to point B. This configuration indicates that the three branches are connected in parallel between A and B. Therefore, the equivalent capacitance between A and B is the sum of the capacitances of the three individual branches. Due to the symmetry of the diagram, it is assumed that all three branches have the same capacitance, say $C$. Thus, the equivalent capacitance $C_{eq} = C + C + C = 3C$.

The diagram shows that the parallel conductors in each branch are separated by a distance $d$. This suggests that each branch acts as a capacitive element, and its capacitance depends on the separation $d$. For a parallel plate capacitor, the capacitance is given by $C = \frac{\epsilon_0 A}{d}$, where $\epsilon_0$ is the permittivity of free space and $A$ is the area of the plates. Assuming that each parallel branch can be modeled as a capacitor with an effective plate area $A$ and separation $d$, the capacitance of each branch is $C = \frac{\epsilon_0 A}{d}$.

Substituting this expression for $C$ into the equation for the equivalent capacitance, we get: $C_{eq} = 3C = 3 \left(\frac{\epsilon_0 A}{d}\right) = \frac{3\epsilon_0 A}{d}$