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Calculate the amount of work done to dissociate a system of three charges of 1muC, 1muC and -4 muC placed on the vertices of an equilateral triangle of side 10 cm. |
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Answer» Solution :As shown in Fig.,` q_1 = 1muC = 1XX 10^(-6) C, q_2 = 1muC =1 XX 10^(-6) C`, `q_3= -4 muC = 4 xx 10^(-6) C and AB = AC = BC = r= 10 cm = 0.1 m` `:.` Work done to dissociate the system of charges `W = - u = -1/(4piepsi_0)[(q_1q_2)/r_12 + (q_1q_3)/r_13 + (q_2 q_3)/r_23]` `=-(9 xx 10^9)[((1XX10^(-6))xx(1xx10^(-6)))/0.1+((1xx10^(-6))xx(-4 xx10^(-6)))/0.1 + ((1 xx 10^(-6))xx(-4 xx 10^(-6)))/0.1]` `=-(9 xx 10^9) [1 xx 10^(-11) - 4 xx 10^(-11) - 4 xx 10^(-11) ] = 0.63 J` 
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