Question

Charge $q_1$ = + 6.0 nC is on Y- axis at y=+3 cm and charge $q_2$=-6.0 nC is on Y- axis at y=-3 cm calculate force on a test charge $q_0$ = 2nC placed on X-axis at x=4 cm.

• A. -51.8 $\hat{i}$ $\mu$N
• B. +51.8 $\hat{i}$ $\mu$N

Answer 1

The net force on $q_0$ is approximately $51.8\,\mu N$ downward (i.e. along $-\hat{j}$); none of the given options are correct.

Answer 2

Solution

We are given

$$q_1=+6.0\,\text{nC}\; \text{at}\;(0, +0.03\,\text{m}),\quad q_2=-6.0\,\text{nC}\; \text{at}\;(0, -0.03\,\text{m}),\quad q_0=+2.0\,\text{nC}\; \text{at}\;(0.04\,\text{m}, 0).$$

For any two charges the Coulomb force is

$$F=k\,\frac{|q_a\,q_b|}{r^2}\quad\text{with }k=9\times10^9\,\text{N m}^2/\text{C}^2.$$

Step 1. Compute distance:
Distance from either $q_1$ or $q_2$ to $q_0$

$$r=\sqrt{(0.04)^2+(0.03)^2}=0.05\,\text{m}.$$

Step 2. Magnitude of force from each charge:

$$F=\frac{9\times10^9\times(2\times10^{-9})(6\times10^{-9})}{(0.05)^2} =\frac{9\times10^9\times12\times10^{-18}}{0.0025} =\;43.2\,\mu\text{N}.$$

Step 3. Directions

For $q_1$ (positive):
Since like charges repel, the force on $q_0$ is along

$$\vec{r_{0}}-\vec{r_{1}}=(0.04-0,\,0-0.03)=(0.04,\,-0.03).$$

Its unit vector is

$$\left(\frac{0.04}{0.05},\,\frac{-0.03}{0.05}\right)=(0.8,\,-0.6).$$

Thus,

$$\vec{F_1}=43.2\,(0.8\,\hat{i}-0.6\,\hat{j})=(34.56\,\hat{i}-25.92\,\hat{j})\;\mu\text{N}.$$

For $q_2$ (negative):
Since opposite charges attract, the force on $q_0$ is directed toward $q_2$. That is,

$$\vec{r_{2}}-\vec{r_{0}}=(0-0.04,\,-0.03-0)=(-0.04,\,-0.03),$$

with unit vector

$$(-0.8,-0.6).$$

Thus,

$$\vec{F_2}=43.2\,(-0.8\,\hat{i}-0.6\,\hat{j})=(-34.56\,\hat{i}-25.92\,\hat{j})\;\mu\text{N}.$$

Step 4. Net force

$$\vec{F}=\vec{F_1}+\vec{F_2}=(34.56-34.56)\,\hat{i}+(-25.92-25.92)\,\hat{j}=(0\,\hat{i}, -51.84\,\hat{j})\;\mu\text{N}.$$

Thus, the net force is about $51.8\,\mu \text{N}$ downward along the negative $y$–axis.

Note on the Given Options

The options provided are

1. $-51.8\,\hat{i}\,\mu N$
2. $+51.8\,\hat{i}\,\mu N$

Both options suggest a force in the $x$–direction. However, our result shows the force is purely vertical (along $-\hat{j}$). Hence, neither option is correct.

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Final Answer:
The net force on $q_0$ is approximately $51.8\,\mu N$ downward (i.e. along $-\hat{j}$); none of the given options are correct.

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Summary of Details:

• Explanation (Minimal):

  1. Calculate the distance $r=0.05\,m$ from each charge to the test charge.
  2. Find force magnitude: $F=43.2\,\mu N$ from each charge using Coulomb’s law.
  3. For $q_1$ the force is repulsive (directed along $(0.04, -0.03)$ → unit vector $(0.8,-0.6)$); for $q_2$ the force is attractive (directed along $(-0.04,-0.03)$ → unit vector $(-0.8,-0.6)$).
  4. Adding forces gives: $\vec{F}=(34.56-34.56)\,\hat{i}+(-25.92-25.92)\,\hat{j}=(0,-51.84)\,\mu N$.

• Answer: The net force is $51.8\,\mu N$ downward (along $-\hat{j}$); hence, none of the provided options [(1) and (2)] are correct.