1. **Problem statement:** Calculate the force experienced by the +5 nC charge due to the other two charges (-9 nC and +4 nC) placed at the vertices of an equilateral triangle with side length 15 cm. 2. **Formula used:** Coulomb's law for force between two point charges: $$F = k \frac{|q_1 q_2|}{r^2}$$ where $k = 8.99 \times 10^9$ Nm$^2$/C$^2$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. 3. **Convert units:** - Charges: $5$ nC = $5 \times 10^{-9}$ C, $-9$ nC = $-9 \times 10^{-9}$ C, $4$ nC = $4 \times 10^{-9}$ C - Distance: $15$ cm = $0.15$ m 4. **Calculate forces on +5 nC charge:** - Force due to $-9$ nC charge: $$F_1 = 8.99 \times 10^9 \times \frac{|5 \times 10^{-9} \times (-9) \times 10^{-9}|}{0.15^2} = 8.99 \times 10^9 \times \frac{45 \times 10^{-18}}{0.0225}$$ - Simplify numerator and denominator: $$= 8.99 \times 10^9 \times 2 \times 10^{-15} = 1.798 \times 10^{-5} \text{ N}$$ - Force due to $+4$ nC charge: $$F_2 = 8.99 \times 10^9 \times \frac{|5 \times 10^{-9} \times 4 \times 10^{-9}|}{0.15^2} = 8.99 \times 10^9 \times \frac{20 \times 10^{-18}}{0.0225}$$ - Simplify numerator and denominator: $$= 8.99 \times 10^9 \times 8.89 \times 10^{-16} = 7.99 \times 10^{-6} \text{ N}$$ 5. **Determine directions:** - The force $F_1$ is attractive (since charges are opposite), directed from +5 nC towards -9 nC. - The force $F_2$ is repulsive (since charges are both positive), directed away from +4 nC. 6. **Calculate resultant force:** - The angle between the two forces is $60^\circ$ (equilateral triangle). - Use law of cosines for magnitude: $$F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 60^\circ}$$ - Substitute values: $$F = \sqrt{(1.798 \times 10^{-5})^2 + (7.99 \times 10^{-6})^2 + 2 \times 1.798 \times 10^{-5} \times 7.99 \times 10^{-6} \times 0.5}$$ - Calculate inside the root: $$= \sqrt{3.23 \times 10^{-10} + 6.38 \times 10^{-11} + 1.44 \times 10^{-10}} = \sqrt{5.31 \times 10^{-10}} = 2.3 \times 10^{-5} \text{ N}$$ 7. **Result:** The force experienced by the +5 nC charge due to the other two charges is approximately $$2.3 \times 10^{-5}$$ newtons.