## Answers

r = AP = distance of charge q_{1} from P = sqrt(8^{2} + 6^{2}) = 10 cm = 0.10 m

a = BP = distance of charge q_{2} from P = 6 cm = 0.06 m

q_{1} = magnitude of charge = 5.30 x 10-6 C

q_{2} = magnitude of charge = 2.35 x 10^{-6} C

E_{1} = electric field by charge q_{1} at P = k q_{1} /r^{2} = (9 x 10^{9}) (5.30 x 10^{-6})/(0.10)^{2}= 4.77 x 10^{6} N/C

E_{2} = electric field by charge q_{2} at P = k q_{2} /a^{2} = (9 x 10^{9}) (2.35 x 10^{-6})/(0.06)^{2}= 5.875 x 10^{6} N/C

= tan^{-1}(8/6) = tan^{-1}(1.33) = 53.06 deg

Components E_{1} Sin along the Y-direction being equal and opposite cancel out . hence the magnitude of net electric field along the X-direction is given as

E = 2 E_{1} Cos + E_{2}

E = 2 (4.77 x 10^{6}) Cos53.06 + (5.875 x 10^{6})

E = 1.16 x 10^{7} N/C

Part B)

direction of net electric field comes out to be

towards the charge q_{2}