Question

two charges 5u(meu)C and -20uC are placed thirty cm apart in the air.the strength of electric field at themid point between them is

Answer 1

Charges are 5μC and -20μC.
Distance between them = 30cm =0.3 m
r = 0.3/2 = 0.15 m
Strength of field at midpoint = $\frac{1}{4 \pi e}$ [ $\frac{( q_{1} )}{ r^{2} }$ + $\frac{ q_{2} }{ r^{2} }$]

                         =  $\frac{1}{4 \pi e}$ [ $\frac{( 5*10^{-6} )}{ 0.15^{2} }$ - $\frac{ 20*10^{-6} }{ 0.15^{2} }$] 

                         = (9×10^9)  [$\frac{( 5*10^{-6} )}{ 0.15^{2} }$ - $\frac{ 20*10^{-6} }{ 0.15^{2} }$]

                         = $-6*10^{6} N$