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A charge q(1) = 2.00 mu C is located at ...

A charge `q_(1) = 2.00 mu C` is located at the origin , and a charge `q_(2) = - 6.00 mu C ` is located at `( 0,3.00 ) m`.
(a) Find the total electric potential due to these charges at the point `P` hose coordinates are ( 4.00,0) m .
( b) Find the change in potential energy of the system of two charges plus a charge `q_(3)= 3.00 muC` as the latter charge moves from infinity to point P .

Text Solution

Verified by Experts

For two charges ,
`V_(P)=k_(e)(q_(1)/(r_(1))+(q_(2))/(r_(2)))`
`V_(P)=(9xx10^(9)Nm^(2)//C^(2))xx((2.00xx10^(-6)C)/(4.00m)-(6.00xx10^(-8)C)/(5.00m))`
` =- 6.3xx10^(3)` V
(b ) When the charge `q_(3) ` is at infinity let us define `U_(i)=0` for the system and when the charge is at P `U_(r)= q_(3)V_(p)` , therefore ,
`Delta=q_(3)V_(P)-0= (3.00xx10^(-6)C)(-6.3xx10^(3)V)`
`=-1.89xx10^(-2)` J
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