Question

For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis in terms of the electric dipole moment, p. what is the magnitude when x equals d or is smaller than d? what is the magnitude when x is greater than d? to view picture: http://www.physicsforums.com/showthread.php?t=425875

Answer 1

by symmetry the field along the x-axis cancels, leaving the vertical component \(\sin\theta\) and since each particle provides its own field the strength of the field is double\[\vec E=k\frac{2q}{r^2}\sin\theta\hat x=k\frac{2p}{dr^2}\sin\theta\hat x\]sine is given by opposite over hypotenuse. The hypotenuse is\[r=\sqrt{(\frac d2)^2+x^2}\]so\[\sin\theta={\frac d2\over\sqrt{(\frac d2)^2+x^2}}=\frac d{2\sqrt{(\frac d2)^2+x^2}}\]plugging in for \(r\) and \(\sin\theta\) gives\[\vec E=k\frac{2p}{d((\frac d2)^2+x^2)}\cdot\frac d{2\sqrt{(\frac d2)^2+x^2}}\hat x=\frac{kp}{((\frac d2)^2+x^2)^{3/2}}\hat x\]

Answer 2

Hi Thank you so much for the answer. i understand exactly what you did. but i dont understand this part of the question: what is the magnitude when x equals d or is smaller than d? what is the magnitude when x is greater than d?