Asked by Danielle
                A charge of - 0.15 C is moved from a position where the electric potential is 13 V to a position where the electric potential is 60 V. What is the change in potential energy of the charge associated with this change in position? 
            
            
        Answers
                    Answered by
            Elena
            
    ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} =
=-7.05 J
    
=-7.05 J
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