Page:Elementary Text-book of Physics (Anthony, 1897).djvu/78

64 $$P,$$ the force acting on it is $$\frac{m}{OP^2};$$ after it has moTed through the infinitesimal distance PB, the force acting on it at $$R$$ is $$\frac{m}{OR^2}\cdot$$ The work done upon it during this motion, equal to the product of the force and the line $$PQ,$$ the projection of its path upon the direction of the force, is greater than $$\frac{m\,.\,PQ}{OP^2}$$ and less than $$\frac{m\,.\,PQ}{OR^2};$$ it may be shown that the work done during this displacement may be represented by $$\frac{m\,. \, PQ}{OP \,. \, OR}\cdot$$ Now $$PQ = OP - OQ = OP - OR$$ in the limit, so that the work done in this displacement equals $$\frac{m(OP - OR)}{OP \,. \, OR} = m\left(\frac{1}{OR} - \frac{1}{OP}\right)\cdot$$ The work done in traversing the following elements of the path, $$BS, ST,$$ etc., is expressed by $$m\left(\frac{1}{OS} - \frac{1}{OR}\right),\,m\left(\frac{1}{OT} - \frac{1}{OS}\right),$$ etc. The work done in traversing the whole path from $$P$$ to $$X$$ is the sum of these expressions, or $$m\left(\frac{1}{OX} - \frac{1}{OP}\right)\cdot$$ By the definition of difference of potential, this expression is equal to the difference of potential between the points $$P$$ and $$X,$$ due to the force of which the centre is $$O$$. If the point $$X$$ lie at an infinite distance from $$O$$, the work done by the force in moving the unit mass to that point equals $$- \frac{m}{OP}\cdot$$ This expression is called the potential at the point $$P$$. It has been obtained on the supposition that the force at $$P$$ is directed toward $$O$$, or is an attractive force. In this case the test unit at an infinite distance possesses the potential energy $$B$$. In moving to $$P$$ the forces of the field do upon it the work $$\frac{m}{OP},$$ so that its potential energy at $$P$$ is $$E - \frac{m}{OP}\cdot$$