Question

Consider an electric dipole on the \(x\) -axis and centered at the origin. At a distance \(h\) along the positive \(x\) -axis, the magnitude of electric field due to the electric dipole is given by \(k(2 q d) / h^{3}\). Find a distance perpendicular to the \(x\) -axis and measured from the origin at which the magnitude of the electric field is the same

Short answer

Answer: The distance, y, perpendicular to the x-axis at which the magnitude of the electric field is the same as at the distance h along the positive x-axis is given by \(y = \sqrt{\frac{h^{3}}{d} - d^{2}}\).

Step-by-step solution

Write down the expression for electric field at an arbitrary point in the xy plane

To find the electric field at an arbitrary point (x, y) in the xy plane due to the electric dipole, we need to consider the electric field contribution from both the positive charge q at position (-d, 0) and the negative charge -q at position (d, 0). The electric field due to a point charge is given by \(kQ / r^{2}\), where Q is the charge, r is the distance from the charge, and k is the electrostatic constant. For the positive charge, the distance from the charge to the point (x, y) is \(\sqrt{(x+d)^{2} + y^{2}}\). Similarly, for the negative charge, the distance from the charge to the point (x, y) is \(\sqrt{(x-d)^{2} + y^{2}}\). Using the formula for the electric field due to a point charge, we can write down the total electric field magnitude \(E\) at the point (x, y) as: \(E = \sqrt{\left(\frac{kq}{(x+d)^{2} + y^{2}} - \frac{kq}{(x-d)^{2} + y^{2}}\right)^{2} + \left(\frac{kqy}{\left((x+d)^{2}+y^{2}\right)^{3/2}} + \frac{kqy}{\left((x-d)^{2}+y^{2}\right)^{3/2}}\right)^{2}}\).

Equate the expressions for electric field magnitudes

We are given that the magnitude of the electric field at distance h along the positive x-axis, i.e., at the point (h, 0) is given by \(k(2 q d) / h^{3}\). We need to find the distance y along the y-axis (perpendicular to the x-axis) from the origin, such that the electric field magnitude at the point (0, y) is the same. To do this, we substitute the values (h, 0) and (0, y) in the electric field magnitude expression from Step 1: \(\frac{k(2 q d)}{h^{3}} = \sqrt{\left(\frac{kq}{(d)^{2} + y^{2}} + \frac{kq}{(d)^{2} + y^{2}}\right)^{2} + \left(\frac{kqy}{\left((d)^{2}+y^{2}\right)^{3/2}} - \frac{kqy}{\left((d)^{2}+y^{2}\right)^{3/2}}\right)^{2}}\). By observing the right-hand side of the equation, we can clearly see that the second term inside the square root will be zero, as the two terms inside the brackets are equal and have opposite signs. Thus, the equation simplifies to: \(\frac{k(2 q d)}{h^{3}} = \sqrt{\left(\frac{2kq}{(d)^{2} + y^{2}}\right)^{2}}\).

Solve for the distance y

Now we can solve the equation obtained in Step 2 for the distance y. \(\frac{k(2 q d)}{h^{3}} = \frac{2kq}{(d)^{2} + y^{2}}\). Divide both sides by \(2kq\): \(\frac{d}{h^{3}} = \frac{1}{(d)^{2} + y^{2}}\). Multiply both sides by \((d)^{2} + y^{2}\): \(d(d^{2} + y^{2}) = h^{3}\). Now, we can solve for y: \(y^{2} = \frac{h^{3}}{d} - d^{2}\). Finally, taking the square root of both sides: \(y = \sqrt{\frac{h^{3}}{d} - d^{2}}\). This is the distance perpendicular to the x-axis and measured from the origin at which the magnitude of the electric field is the same as at the distance h along the positive x-axis.