A thundercloud may be idealized as a vertical dipole with its positive charge uppermost. Consider the changes in potential gradient at a point \(\mathrm{A}\) on the earth's surface as such a cloud approaches at a constant height from a great distance to a point directly over A. Assume that the earth's surface can be treated as a plane conductor. For what horizontal distance \(R\) is the ficid at A zero if the height of the cloud is \(H\) ?

For the vertical dipole, assume there is a positive charge \(q\) at height \(H\), and a negative charge \(-q\) at the bottom. As we are asked to find the potential, let's first establish the potential formula: \[V = k \dfrac{q}{r}\] Where: \(V\) is the electric potential, \(k\) is an electrostatic constant, \(k = 8.9875 × 10^9 N·m^2/C^2\), \(q\) is the point charge, \(r\) is the distance from the charge to the point. The potential at point A due to the positive charge is: \[V_+ = k \dfrac{q}{\sqrt{H^2 + R^2}}\] And, the potential at point A due to the negative charge is: \[V_- = k \dfrac{-q}{\sqrt{H^2 + X^2}}\] Where, \(X\) is the horizontal distance between the negative charge and point A. Now, we need to find the potential at point A due to both charges. As the charges have opposite signs, the potentials are added: \[V_A = V_+ + V_-\]

To find the electric field at A, we need to compute the gradient of the potential at A. The gradient can be represented as \(\nabla V_A\) : \[\nabla V_A = \dfrac {\partial V_A}{\partial X}\] Substitute the expression of \(V_A\): \[\nabla V_A = \dfrac{\partial}{\partial X} \left( k \dfrac{q}{\sqrt{H^2 + R^2}} + k \dfrac{-q}{\sqrt{H^2 + X^2}} \right)\] Use the chain rule of differentiation, the gradient becomes: \[\nabla V_A = kq \left( \dfrac{-R}{(H^2 + R^2)^{3/2}} + \dfrac{-X}{(H^2 + X^2)^{3/2}} \right)\]

We are asked to find the distance R where the field at A is zero. To do so, set \(\nabla V_A\) to zero: \[0 = kq \left( \dfrac{-R}{(H^2 + R^2)^{3/2}} + \dfrac{-X}{(H^2 + X^2)^{3/2}} \right)\] \[0 = \dfrac{-R}{(H^2 + R^2)^{3/2}} + \dfrac{-X}{(H^2 + X^2)^{3/2}}\] Now, rearrange the equation to find R: \[R = X \dfrac{\sqrt{(H^2 + X^2)^3}}{\sqrt{H^2 + R^2}}\] \[R = X \left(\dfrac{H^2+X^2}{H^2+R^2}\right)^{1/2}\] This equation represents the relationship between the distance \(R\) (where the electric field at A is zero) and the horizontal distance X from A to the negative charge of the dipole.