Two point charges 3.0 cm apart have an electric potentialenergy -150 J. The tota

Question

Two point charges 3.0 cm apart have an electric potentialenergy -150 J. The total charge is 40 nC. What is the lesser charge? What is thelarger charge? I made a change from 3.0 mm to 3.0cm how does that change the problem Two point charges 3.0 cm apart have an electric potentialenergy -150 J. The total charge is 40 nC. What is the lesser charge? What is thelarger charge? I made a change from 3.0 mm to 3.0cm how does that change the problem What is the lesser charge? What is thelarger charge? I made a change from 3.0 mm to 3.0cm how does that change the problem

Explanation / Answer

   ElectrostaticP.E.   U   =   -(1/40) * q1 * q2 /r    here   q1   +   q2   =   40nC   =   40 *10-9   C   =>   q2   =  ( 40- q1) * 10-9   C    - 150 *10-6   =   - 9.0 *109 * q1 * ( 40 - q1) *10-18/ 3.0 * 10-3    150 * 10-6 / 3.0 *10-6   =   40 *q1   -   q12    q12   -   40*q1   +   50   =   0    Solving aboveequation      q1   =   1.29   nC    q2   =   38.71   nC       q12   -   40*q1   +   50   =   0    Solving aboveequation      q1   =   1.29   nC    q2   =   38.71   nC   

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