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Two point charges, q1 at position (2, 2, 1) and q2 at the origin, exert an electrostatic force on each other. Calculate the force on q2 due to q1.

Physics

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Answer

(3)\text{(3)}

Step-by-step solution

5 steps

  1. Identify the given positions of the charges.

    The position of charge q1q_1 is r1⃗=(2i^+2j^+k^)\vec{r_1} = (2\hat{i} + 2\hat{j} + \hat{k}).
    The position of charge q2q_2 is r2⃗=(0i^+0j^+0k^)\vec{r_2} = (0\hat{i} + 0\hat{j} + 0\hat{k}).

  2. Calculate the displacement vector from q1q_1 to q2q_2.

    The displacement vector r⃗21=r2⃗−r1⃗\vec{r}_{21} = \vec{r_2} - \vec{r_1} is:

    r⃗21=(0i^+0j^+0k^)−(2i^+2j^+k^)\vec{r}_{21} = (0\hat{i} + 0\hat{j} + 0\hat{k}) - (2\hat{i} + 2\hat{j} + \hat{k}) r⃗21=−2i^−2j^−k^\vec{r}_{21} = -2\hat{i} - 2\hat{j} - \hat{k}

  3. Calculate the magnitude of the displacement vector.

    The magnitude ∣r⃗21∣|\vec{r}_{21}| is:

    ∣r⃗21∣=(−2)2+(−2)2+(−1)2|\vec{r}_{21}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} ∣r⃗21∣=4+4+1|\vec{r}_{21}| = \sqrt{4 + 4 + 1} ∣r⃗21∣=9|\vec{r}_{21}| = \sqrt{9} ∣r⃗21∣=3|\vec{r}_{21}| = 3

  4. Apply Coulomb's Law to find the electrostatic force.

    The electrostatic force F⃗\vec{F} on charge q2q_2 due to charge q1q_1 is given by:

    F⃗=14πϵ0q1q2∣r⃗21∣3r⃗21\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}

    Substitute the calculated values:

    F⃗=14πϵ0q1q2(3)3(−2i^−2j^−k^)\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{(3)^3} (-2\hat{i} - 2\hat{j} - \hat{k}) F⃗=14πϵ0q1q227(−2i^−2j^−k^)\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{27} (-2\hat{i} - 2\hat{j} - \hat{k}) F⃗=q1q2108πϵ0(−2i^−2j^−k^)\vec{F} = \frac{q_1 q_2}{108\pi\epsilon_0} (-2\hat{i} - 2\hat{j} - \hat{k})

    This can be rewritten by factoring out a negative sign:

    F⃗=−q1q2108πϵ0(2i^+2j^+k^)\vec{F} = -\frac{q_1 q_2}{108\pi\epsilon_0} (2\hat{i} + 2\hat{j} + \hat{k})

  5. Compare the result with the given options.

    The calculated force matches option (3).

    The final answer is (3)\text{(3)}.

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