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Suppose that three point charges q_(a), q_(b) and q_(c) are arranged at the vertices of a right -angled triangle as shown . What is the absolute electric potential at the position of the third charge if q_(a)=- 6.0 mu C , q_(b)=+4.0muC , q_(c)=+2.0 mu C , a= 4.0 mand b=3.0 m ? Suppose that the third charge which is initially at rest is repelled to infinity by the combined electric field of the other two charges which are held fixed . What is the final kinetic energy of the third charge ? |
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Answer» Solution :The electric POTENTIAL at P (the position of the third charge ) due to the presence of the first charge is `V_(a)=K_(e)(q_(a))/(c)=(9XX10^(9))((-6xx10^(-6)))/((sqrt(4^(2))+3^(2)))=-1.08xx10^(4)`V Likewise the electric potential due to the presence of the second charge is `V_(b)=K_(e)(q_(b))/(b)=(9xx10^(9))((4xx10^(-6)))/((3))=1.20xx10^(4)`V. The net potential of the third charge `V_(c)` is simply the algebraic sum of the potentials due to the OTHERS two CHARGES taken in isolation . Thus `V_(c)=V_(a)+V_(b)=1.20xx10^(3)V` |
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