Answer
a) $x=0.18m$
b) $8 cm$ from $-2\mu C$ from left side
$1.6 cm$ from $-2\mu C$ from right side
Work Step by Step
a) $E_1=E_2$
$\frac{kq_1}{r_1^2}=\frac{kq_2}{r_2^2}$
For the electric field to equal 0, the magnitude of the electric field due to each charge needs to be the same but in different directions. This can only be a certain distance x from the negative charge on the opposite side from the positive charge. If it was in between the two charges, the electric field will be in the same direction for both charges, so the sum cannot be 0. If it is on the same side as the positive charge, the magnitude of the electric field from the negative charge will never equal the magnitude of the electric field from the positive charge because it is farther and has a smaller charge magnitude.
$x=\frac{d\sqrt{q_2}}{\sqrt{q_1}-\sqrt{q_2}}$
$x=\frac{(0.04m)\sqrt{2.0\mu C}}{\sqrt{3.0\mu C}-\sqrt{2.0\mu C}}=0.18m$
b) $\frac{kq_1}{d-x_1}=\frac{kq_2}{x_1}$
$q_1x_1=q_2(d-x_1)$
$(3\mu C)x_1=(-2\mu C)(0.04m-x_1)$
$(1\mu C)x_1=(-2\mu C)0.04m$
$x_1=-0.08m$, $8 cm$ from $-2\mu C$ from left side
$\frac{kq_1}{d+x_1}=\frac{kq_2}{x_1}$
$q_1x_1=q_2(d+x_1)$
$(3\mu C)x_1=(-2\mu C)(0.04m+x_1)$
$(5\mu C)x_1=(-2\mu C)0.04m$
$x_1=-0.0016m$, $1.6 cm$ from $-2\mu C$ from right side
