Consider the particle with the small positive chargelocated at some distance from a positively charged sphere inFigure 22.24b. If you push the particle closer to the sphere,you will expend energy to overcome electrical repulsion; thatis, you will do work in pushing the charged particle againstthe electric field of the sphere. This work done in moving theparticle to its new location increases its energy. We call theenergy the particle possesses by virtue of its location,electric potential energy. If the particle is released, itaccelerates in a direction away from the sphere, and itselectric potential energy changes to kinetic energy.FIGURE 22.24What will happen to the electrical potential energy and work If we instead push aparticle with twice the charge?Calculate the electric potential energy and work, given that the radial distancebetween two charges at point A is 0.150 meters and point B is 0.110 meters. The sphere(q) has a charge of +4.30 X 10* Coulomb.Charge, q.(C,Coulomb)Electric Potential Energy,EPEWork ((J,EPE + Charge = U/q,(V, Volts)Joule)(J, Joule)UA:Us:AU(using U.) (using Us)Using AU2.40 X 10*UA:Us:AU4.80 X 10Ug:AU9.6 X 10*Equations:U, = k q, q/r.U. = k q, q/r.W = -AU = -(U, - U) = U, - U,„U = U. - U,

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