Electrostatics

JEE Main level, single correct, medium

6 of 311

Question 6 of 311

Two point charges q1=3 μCq_1=3\,\mu C and q2=−4 μCq_2=-4\,\mu C are placed at points (2i^+3j^+3k^)(2\hat{i}+3\hat{j}+3\hat{k}) and (i^+j^+k^)(\hat{i}+\hat{j}+\hat{k}) respectively. Force on charge q2q_2 is ________ N. (Take 14πϵ0=9×109 SI Units)\left(\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times 10^9 \text{ SI Units}\right)

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Answer: (B) (4i^+8j^+8k^)×10−3(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}

Position vector of q1q_1, r⃗1=2i^+3j^+3k^\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}

Position vector of q2q_2, r⃗2=i^+j^+k^\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}

Vector from q1q_1 to q2q_2 is r⃗21=r⃗2−r⃗1=(i^+j^+k^)−(2i^+3j^+3k^)=−i^−2j^−2k^\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}

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

Force on q2q_2 due to q1q_1 is given by Coulomb's law in vector form:

F⃗21=14πϵ0q1q2∣r⃗21∣3r⃗21\vec{F}_{21} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}

Substituting the given values:

F⃗21=9×109×(3×10−6)×(−4×10−6)33(−i^−2j^−2k^)\vec{F}_{21} = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (-4 \times 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})

F⃗21=−108×10−327(−i^−2j^−2k^)\vec{F}_{21} = \dfrac{-108 \times 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})

F⃗21=−4×10−3(−i^−2j^−2k^)\vec{F}_{21} = -4 \times 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k})

F⃗21=(4i^+8j^+8k^)×10−3 N\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N}

Answer: (4i^+8j^+8k^)×10−3(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}

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