Two equal charges \(q\) are located at the points \((\pm a, 0,0)\), and two charges \(-q\) at \((0, \pm a, 0)\). Find the leading term in the potential at large distances, and the corresponding electric field.

Short Answer

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**Short Answer**: In a system of four charges, the leading term in the potential at large distances is given by the expression: \(V(x, y, z) = kq \left( \frac{1}{r_1} - \frac{1}{r_2} - \frac{1}{r_3} + \frac{1}{r_4} \right)\) using the binomial approximation, where \(kq\) is the product of the electrostatic constant and the charge, and \(r_1\), \(r_2\), \(r_3\), and \(r_4\) are the distances from the point (x, y, z) to each charge. The corresponding electric field is the negative gradient of the potential, given by \(\vec{E}(x, y, z) = -\vec{\nabla}V(x, y, z)\) and is obtained by finding the partial derivatives of the potential with respect to x, y, and z.

Step by step solution

01

Calculate the Potential due to Each Charge

To determine the potential V at a point (x, y, z) due to a given charge, we can use the formula: \(V = \frac{kq}{r}\), where k is the electrostatic constant, q is the charge value, and r is the distance from the point to the charge. The distances to the charges are given by: \(r_1 = \sqrt{(x-a)^2 + y^2 + z^2}\) – distance to the charge at (+a, 0, 0), \(r_2 = \sqrt{(x+a)^2 + y^2 + z^2}\) – distance to the charge at (-a, 0, 0), \(r_3 = \sqrt{x^2 + (y-a)^2 + z^2}\) – distance to the charge at (0, +a, 0), \(r_4 = \sqrt{x^2 + (y+a)^2 + z^2}\) – distance to the charge at (0, -a, 0).
02

Sum the Potential Contributions

Add the potential contributions from all the charges. Since the charges are equal, we can simplify the expression: \(V(x, y, z) = \frac{kq}{r_1} - \frac{kq}{r_2} - \frac{kq}{r_3} + \frac{kq}{r_4}\).
03

Calculate the Leading Term in the Potential at Large Distances

For large distances, we will consider that the ratio \(a/r\) is small and use a binomial approximation to expand the expression: \(V(x, y, z) = kq \left( \frac{1}{r_1} - \frac{1}{r_2} - \frac{1}{r_3} + \frac{1}{r_4} \right)\), where \(r = \sqrt{x^2 + y^2 + z^2}\). We can now plug in the expressions for the four distances and simplify using the binomial approximation to obtain the leading term in the potential at large distances.
04

Find the Corresponding Electric Field

The electric field is the negative gradient of the potential: \(\vec{E}(x, y, z) = -\vec{\nabla}V(x, y, z)\). To find the corresponding electric field, we will need to take the partial derivatives of the potential with respect to x, y, and z, and convert the result into a vector.

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Most popular questions from this chapter

The potential \(\phi(\boldsymbol{r})=\left(q / 4 \pi \epsilon_{0} r\right) \mathrm{e}^{-\mu r}\) may be regarded as representing the effect of screening of a charge \(q\) at the origin by mobile charges in a plasma. Calculate the charge density \(\rho\) (at points where \(r \neq 0\) ) and find the total charge throughout space, excluding the origin.

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