A+250 esu of charge P is On the line joining other two charges `Q` (`+50` esu of charge ) and `R(-300` esu of charge) in between them. Distance of P from Q is 5 cm and from R it is 10 cm. What the resultant force acting on P?
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Explanation

The force actging on P due to the charge Q
`=(50xx250)/((5)^(2))=500` dyn, along `vec(PR)`
The force acting on P due to the charge R
`=(250xx300)/((10)^(2))=750` dyn, along `vec(PR)`
`:gt` The resultant force acting one the charge P
`=500+750=1250` dyn, along `vec(PR)`