Question

The figure shows an arrangement of four charged particles, with angle $θ = 35.0 °$ and distance $d = 2.00 cm$ . Particle $2$ has charge $q_{2}$ $= + 8.00 \times 10^{- 19} C$ , particles $3$ and $4$ have charges $q_{3} = q_{4} = - 1.60 \times 10^{- 19} C$ .
$(a)$ What is distance $D$ between the origin and particle $2$ if the net electrostatic force on particle $1$ due to the other particles is zero?

$(b)$ If particles $3$ and $4$ were moved closer to the $x$ -axis but maintained their symmetry about that axis, would the required value of $D$ be greater than, less than, or the same as in part $(a)$ ?

Accepted Answer

(a) We note that $\cos (35 °) = \frac{1}{2} \sqrt{2.7}$ , so that the dashed line distance in the figure is
$r = \frac{2 d}{\sqrt{2.7}}$ .

The net force on $q_{1}$ due to the two charges $q_{3}$ and $q_{4}$ (with $|q_{3}| = |q_{4}| = 1.60 \times$ $10^{- 19} C)$ on the $y$ -axis has magnitude

$2 \frac{|q_{1} q_{3}|}{4 π ε_{0} r^{2}} \cos (35 °) = \frac{2.7 \sqrt{2.7} |q_{1} q_{3}|}{16 π ε_{0} d^{2}}$ .

This must be set equal to the magnitude of the force exerted on $q_{1}$ by $q_{2} = 8.00 \times 10^{- 19} C$ $= 5.00 |q_{3}|$ in order that its net force be zero,

$\frac{2.7 \sqrt{2.7} |q_{1} q_{3}|}{16 π ε_{0} d^{2}} = \frac{|q_{1} q_{2}|}{4 π ε_{0} (D + d)^{2}} \Rightarrow D = d (2 \sqrt{\frac{5}{2.7 \sqrt{2.7}}} - 1) = 1.25 d$ .

Given, $d = 2.00 cm$ . This, then leads to $D = 2.50 cm$ .

(b) As the angle decreases, its cosine increases, resulting in a larger contribution from the charges on the $y$ -axis.

To offset this, the force exerted by $q_{2}$ must be made stronger, so that it must be brought closer to $q_{1}$ (keep in mind that Coulomb's law is inversely $y$ proportional to distance-squared). Thus, $D$ must be decreased.

Important Questions on Coulomb's Law

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