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Suppose that the same system of charges is now placed in an external electric field E=A(1//r^2),A=9xx10^5Cm^-2. What would the electros -tatic energy of the configuration be ? |
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Answer» Solution :The mutal INTERACTION energy of the TWO charges REMAINS unchanged. In addition , there is the energy of interaction of the two charges with the external electric FIELD. We find, `q_(1)V(r_1)+q_2V(r_2)=A(7muC)/(0.09m)+A(-2muC)/(0.09m)` and the net electrostatic energy is `q_1V(r_1)+q_2V(r_2)+(q_1q_2)/(4piepsi_0r_12)` `=A(7muC)/(0.09m)+A(-2muC)/(0.09m)-0.7J` `=70-20-0.7=49.3J`. |
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