Question

Point charges are located at 3, 8, and 11 cm along the x-axis(+q, -2q, +q). What is the x-component of the force on the charge located at x=8 cm given that q=1.15nc?

Answer 1

Approximately
`1.69 * 10^(-23)\; {\rm N}`.

Step-by-step explanation:

Look up the value of Coulomb's Constant:
`k \approx 8.988 * 10^(-9)\; {\rm N \cdot m^(2) \cdot C^(-2)}`.

Consider point charges of magnitude
`q_(1)` and
`q_(2)`. If the distance between these charges is
`r`, the magnitude of the electrostatic force between them would be
`(k\, q_(1)\, q_(2)) / (r^(2))`.

In this question, the two
`(+q)` charges are
`5\; {\rm cm}` and
`3\; {\rm cm}` away from the center
`(-2\, q)` charge, respectively. Convert units to standard unit of distance (meters,
`{\rm m}`) and charge (coulombs,
`{\rm C}`):

`q = 1.15 \; {\rm nC} = 1.15 * 10^(-9)\; {\rm C}`.

`\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} * \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}`.

`\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} * \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}`.

The magnitude of the electrostatic forces on the
`(-2\, q)` charge would be:

`\begin{aligned}(k\, q_(1)\, q_(2))/(r^(2)) &\approx \frac{1}{(0.05\; {\rm m})^(2)} \\ &\quad * (8.988 * 10^(-9)\; {\rm N \cdot m^(2) \cdot C^(-2)})\\ &\quad * ((-2) \, (1.15* 10^(-9)\; {\rm C}))\, (1.15* 10^(-9)\; {\rm C})) \\ &\approx 9.509* 10^(-24)\; {\rm N}\end{aligned}`.

`\begin{aligned}(k\, q_(1)\, q_(2))/(r^(2)) &\approx \frac{1}{(0.03\; {\rm m})^(2)} \\ &\quad * (8.988 * 10^(-9)\; {\rm N \cdot m^(2) \cdot C^(-2)})\\ &\quad * ((-2) \, (1.15* 10^(-9)\; {\rm C}))\, (1.15* 10^(-9)\; {\rm C})) \\ &\approx 2.641* 10^(-23)\; {\rm N}\end{aligned}`.

Since the charges are of opposite sign, the
`(-2\, q)` charge would attract both of the
`(+q)` charges. In particular, the (approximately)
`9.509* 10^(-24)\; {\rm N}` force would point to the left. The (approximately)
`2.641 * 10^(-23)\; {\rm N}` force would point to the right.

As a result, the net force on the
`(-2\, q)` charge would point to the right. The magnitude of the net force on this charge would be approximately
`2.641 * 10^(-23)\; {\rm N} - 9.509* 10^(-24)\; {\rm N} \approx 1.69 * 10^(-23)\; {\rm N}`.