Find the magnitude of the resultant force on a charge of ` 1 muC ` held at P due to two charges of `+ 2 xx 10 ^(-8) C and -10^(-8) C ` at A and B , respectively.
Take , AP = 10 cm and BP = 5 cm and ` angle APB = 90^(@) `
` (##ARH_CC_CHSE_ODI_PHY_XII_C01_E01_010_Q01.png" width="80%">

Explanation

Here, charge at P , q= ` 1 mu C = 10 ^(-6)C `
Charge at ` A , q_1 = 2 xx 10 ^(-8) C `
Charge at `B,q_2 = - 10 ^(-8)C `
AP = 10 cm = 0.1 m , BP = 5 cm = 0.05 m,
` angle APB = 90^(@) , F = ? `
Force at P due to `q_1 ` charge at A
` F_1 = ( 1)/(4 piin_0) (q_1q)/(AP^(2)) ` along AP produced
` " " = (9 xx 10 ^(9) xx 2 xx 10^(-8) xx 10 ^(-6))/((0.1)^(2)) = 18 xx 10 ^(-3)N`
Force at P due to `q_2` charge at B,
`F_2 = (1)/(4piin_0) (q_2q)/(BP^(2)) ` along PB produced
` ( 9 xx 10 ^(9) xx - 10 ^(-8) xx 10 ^(-6))/((0.5)^(2)) = - 36 xx 10 ^(-3) N`
As angle between `F_1 and F_2 " is " 90 ^(@) `
Resultant force
` F= sqrt(F_1^(2) + F_2^(2)) = sqrt((18 xx 10 ^(-3))^(2) + (-36 xx 10 ^(-3))^(2)) `
` = sqrt((324 +1296 ) xx 10 ^(-6)) = sqrt(1620 xx 10 ^(-6)) `
` = 40 .2 xx 10 ^(-3)`
` = 4.0 xx 10 ^(-2) N`