Question

What is the magnitude of the electric field at a point midway between a −7.1μC and a +5.0μC charge 8.8cm apart? Assume no other charges are nearby.

Answer 1

Electric field direction is always directed away from positive charge and towards negative charge

Both fields are in same direction, so add individual electric fields

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

$E=kq_{1}/d^{2}+kq_{2}/d^{2}$

Distance is same (midpoint) =4.4cm =0.044 m

$E=(q_{1}+q_{2})k/d^{2}$

$E=(7.1*10^{-6}C+5*10^{-6}C)*9*10^{9}/(0.044m)^{2}$

$E=(7.1*10^{-6}+5*10^{-6})*9*10^{9}/(0.044)^{2}$

$E=12.1*10^{-6}*9*10^{9}/(0.044)^{2}$

$E=108900/(0.044)^{2}$

ANSWER: $E=5.625*10^{^{7}}N/C$

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