A point charge q is placed at the origin. A second point charge -3q is placed at

Question

A point charge q is placed at the origin. A second point charge -3q is placed at the point (7, 0). Find the equation of the equipotential curve in the xy plane where the potential due to these two charges is zero. Please present your answer in the form y equals a function of x, or y^2 equals a function of x. Show that the orbital radius r of a charged particle moving at right angles to a magnetic field B can be written r = Squareroot 2 Km/(qB), where K is the kinetic energy of the particle, m is its mass and q is its charge.

Explanation / Answer

Potental at a point is sum of potential due to all charges.
Potential due to a point charge q at a distance r from, from itself is given by
V = k q /r , where k is a constant.

Let potential due to two given charges be zero, at point (x,y).
Distance of point (x,y) from origin is (x2 + y2)1/2 and from (7,0) is ( (x-7)2 + y2)1/2

Hence potential at (x,y) due to two charges is
kq ( 1/(x2 + y2)1/2 - 3 / ( (x-7)2 + y2)1/2 )
As this potential is zero, term in the brackets is zero. Hence
1/(x2 + y2)1/2 - 3 / ( (x-7)2 + y2)1/2 = 0
3 ((x2 + y2)1/2 = (x-7)2 + y2)1/2
9 (x2 + y2) = (x-7)2 + y2)
y2 = -x2 -14x +49
This is the required equation

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