Understanding Equipotential Curves Around Perpendicular Infinite Rods

The question asks about the shape of equipotential curves created by two infinite straight rods that are placed perpendicular to each other. Equipotential curves are lines or surfaces where the electric potential is constant. To find these curves, we first need to determine the electric potential at any point in the plane due to this charge distribution.

Electric Potential of a Single Infinite Straight Rod

The electric potential due to an infinitely long straight line of charge (like an infinite rod) at a distance \(r\) from the rod is given by a logarithmic function. Assuming the potential is zero at some reference distance \(r_0\), the potential is proportional to \(-\ln(r)\). If we consider the potential relative to infinity, it's typically defined as zero, but for an infinite line charge, we must use a finite reference point or express potential difference. However, the functional form of the potential is key: it depends on the logarithm of the perpendicular distance from the rod.

Total Potential from Two Perpendicular Infinite Rods

Let's assume the two infinite straight rods lie along the x-axis and the y-axis in a 2D plane. Consider a point \((x, y)\) in this plane. The perpendicular distance of this point from the rod along the y-axis is \(|x|\). The perpendicular distance from the rod along the x-axis is \(|y|\).

The electric potential at point \((x, y)\) due to the rod along the y-axis is proportional to \(-\ln(|x|)\). Let's write this as \(V_1 = -k \ln(|x|) + C_1\), where \(k\) is a constant related to the linear charge density and permittivity, and \(C_1\) is an integration constant.

Similarly, the electric potential at point \((x, y)\) due to the rod along the x-axis is proportional to \(-\ln(|y|)\). Let's write this as \(V_2 = -k \ln(|y|) + C_2\).

Since electric potential is a scalar quantity, the total potential \(V\) at point \((x, y)\) due to both rods is the sum of the individual potentials:

$$V(x, y) = V_1 + V_2 = -k \ln(|x|) - k \ln(|y|) + C_1 + C_2$$

Let \(C = C_1 + C_2\) be the total integration constant. So, the total potential is:

$$V(x, y) = -k (\ln(|x|) + \ln(|y|)) + C$$

Using the property of logarithms, \(\ln(a) + \ln(b) = \ln(ab)\), we get:

$$V(x, y) = -k \ln(|x| |y|) + C$$

Equation for Equipotential Curves

For equipotential curves, the potential \(V(x, y)\) is constant. Let this constant potential be \(V_0\). Setting the equation for potential equal to \(V_0\):

$$V_0 = -k \ln(|x| |y|) + C$$

Rearranging the terms:

$$V_0 - C = -k \ln(|x| |y|)$$

$$\frac{C - V_0}{k} = \ln(|x| |y|)$$

Let \(\frac{C - V_0}{k}\) be a constant, say \(A\). Then, \(A = \ln(|x| |y|)\).

Taking the exponential of both sides:

$$e^A = |x| |y|$$

Let \(K = e^A\), where \(K\) is a positive constant. The equation for the equipotential curves is therefore:

$$|xy| = K$$

Identifying the Shape of the Equipotential Curves

The equation \(|xy| = K\) describes curves where the product of the absolute values of the x and y coordinates is a positive constant. This equation represents a hyperbola. Specifically:

• In the first quadrant (\(x > 0, y > 0\)), the equation is \(xy = K\).
• In the second quadrant (\(x < 0, y > 0\)), the equation is \((-x)y = K\), or \(xy = -K\).
• In the third quadrant (\(x < 0, y < 0\)), the equation is \((-x)(-y) = K\), or \(xy = K\).
• In the fourth quadrant (\(x > 0, y < 0\)), the equation is \(x(-y) = K\), or \(xy = -K\).

These curves are graphs of \(y = K/x\) (in quadrants 1 & 3) and \(y = -K/x\) (in quadrants 2 & 4). The asymptotes of these hyperbolas are the x-axis (\(y=0\)) and the y-axis (\(x=0\)), which are the lines where the infinite rods are located. Since the asymptotes are perpendicular, these hyperbolas are known as rectangular hyperbolas.

Thus, the equipotential curves in the plane of two perpendicular infinite straight rods are rectangular hyperbolas. These Equipotential Curves show regions of constant potential.