Question

A +6.0 µC charge (charge A) is placed at the origin (i.e., x = 0 m), a +9.0 µC charge (charge B) is placed at x = −5.0 cm and a −3.0 µC charge (charge C) is placed at x = +2.0 cm.

What is the net force on charge A due to charges B and C? (to 2 s.f and in N)

Answer 1

Explanations & Calculations

• To calculate electrostatic forces, the coulomb equation is used.
• $\frac{1}{4\pi \epsilon} = 9\times 10^9Nm^2C^{-2}$

• Repulsive forces are generated between A & B due to the same polarity and attractive forces between A & C for being opposite polarity.

• Repulsive force on A due to B,

$\qquad\qquad \begin{aligned} \small F &= \small \frac{1}{4\pi \epsilon_0} \frac{(9\times10^{-6}C)(6\times10^{-6}C)}{(0.05m)^2} \\ &= \small 9\times10^9 Nm^2C^{-2} \times 2.16\times10^{-8}m^{-2}C^{2}\\ &= \small 194.4\,N \end{aligned}$

• Attractive force on A due to C,

$\qquad\qquad \begin{aligned} \small \color{red}F &= \small \frac{1}{4\pi \epsilon_0} \frac{(6\times10^{-6}C)(3\times10^{-6}C)}{(0.02m)^2}\\ &= \small \color{red}405\,N \end{aligned}$

• Therefore, net force on A,

$\qquad\qquad \begin{aligned} \small F+\color{red}F &= \small 194.4N + \color{red}405.0N\\ &= \small \bold{599.40N}\\ &= \small \bold{5.9*10^2N} \end{aligned}$