1.

ABCD is a square of side 2m. Charges of 5nC, + 10nC and -5nC are placed at corners A,B and C respectively .What is the workdone in transferring a charge of 5nC from 'D' to the point of intersection of the diagonals ?

Answer»

SOLUTION :
In the Fig. .O. represents the point of intersection of the diagonals.
AO=BO=CO=DO=`sqrt2`
Total electric potential at .O. is
`V_0=1/(4piepsilon_0)[q_1/(AO)+q_2/(BO)+q_3/(CO)]`
`=9xx10^9[(5xx10^(-9))/sqrt2+(10xx10^(-9))/sqrt2-(5xx10^(-9))/sqrt2]`
`=9xx10^9xx10^(-9)[5/sqrt2+10/sqrt2-5/sqrt2]`
=9[3.536+7.072-3.536]
`V_0`=63.648 volt
Total electric potential at D is
`V_D=1/(4pi in_0)[q_1/(AD)+q_2/(BD)+q_3/(CD)]`
`=9xx10^9 [(5xx10^(-9))/2+ (10xx10^(-9))/(2SQRT2)-(5xx10^(-9))/2]`
`=9xx10^9xx10^(-9)[5/2+5/sqrt2-5/2]`
=9[25+3.536-2.5] = 31.824 volt
`therefore` Potential difference between O and D is
`V=V_0-V_D`=63.648-31.824=31.824 volt
`therefore` Work required to transfer a charge of `5xx10^(-9)` C FROMD to O is
`w=Vq=31.824xx5xx10^(-9)`
`w=159.12xx10^(-9)` Joule .


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