Question

In the figure, the net electrostatic force on charge Q_A is zero. If Q_A = +2.83 nC andQ = -1.33 nC, determine Q₀

Answer 1

+Q0 破 0 a QA 3 (+2a, -2a)

From figure above the force of repulstion F1 has a module equal to:

$F_1 = \frac{kQ_0Q_A}{(2\sqrt{2}a)^2}$

the module of F2 is:

$F_2 = \frac{kQQ_A}{(\sqrt{5}a)^2}$

if the net force is zero:

$F_1 =2F_2\cos\theta$

where the angle is:

$\theta = 45 - \alpha$

form the upper right triangle

$\tan \alpha = 1/2 \rightarrow \alpha = 26.6$ so $\theta = 18.4$

so:

$\frac{kQ_0Q_A}{8a^2}=2\left ( \frac{kQQ_A}{5a^2} \right )\cos 18.4$

soving for Q0 we have

$Q_0 = 3.036 \left |Q \right |= 4.038 nC$