What is the force on q₂=−4μC at (1,1,1) due to q₁=3μC at (2,3,3)?

(4î+8ĵ+8k̂)×10⁻³ N. r⃗₁₂=(−1,−2,−2), |r|=3m. F=kq₁q₂r⃗₁₂/r³=(4,8,8)×10⁻³ N.

Why is force in direction (1,2,2) rather than (−1,−2,−2)?

q₁q₂<0 (opposite charges, attractive). The product kq₁q₂<0 combined with r⃗₁₂=(−1,−2,−2) gives F in direction (1,2,2), i.e., toward q₁.

q₁=3μC at (2,3,3) q₂=−4μC at (1,1,1) — force on q₂

Q: Why is force in direction (1,2,2) rather than (−1,−2,−2)?

q₁q₂<0 (opposite charges, attractive). The product kq₁q₂<0 combined with r⃗₁₂=(−1,−2,−2) gives F in direction (1,2,2), i.e., toward q₁.

JEERANKERS

PhysicsElectrostatics

QMCQElectrostatics

Two point charges $q_1=3\,\mu\text{C}$ at $(2\hat{i}+3\hat{j}+3\hat{k})$ m and $q_2=-4\,\mu\text{C}$ at $(\hat{i}+\hat{j}+\hat{k})$ m. Force on $q_2$? ($k=9\times10^9$ SI)

A) $(12\hat{i}+24\hat{j}+24\hat{k})\times10^{-3}$ B) $(4\hat{i}+8\hat{j}+8\hat{k})\times10^{-3}$
C) $(3\hat{i}+6\hat{j}+6\hat{k})\times10^{-3}$ D) $(-4\hat{i}-8\hat{j}-8\hat{k})\times10^{-3}$

✅ Correct Answer

B) (4î+8ĵ+8k̂)×10⁻³ N

✓

Solution

Displacement vector from q₁ to q₂

$\vec{r}_{12}=(1-2)\hat{i}+(1-3)\hat{j}+(1-3)\hat{k}=-\hat{i}-2\hat{j}-2\hat{k}$
$|\vec{r}_{12}|=\sqrt{1+4+4}=3$ m

Apply vector Coulomb's Law

$\vec{F}=\dfrac{kq_1q_2}{|\vec{r}_{12}|^3}\vec{r}_{12}=\dfrac{9\times10^9\times3\times10^{-6}\times(-4\times10^{-6})}{27}(-\hat{i}-2\hat{j}-2\hat{k})$

$=-4\times10^{-3}\times(-\hat{i}-2\hat{j}-2\hat{k})=(4\hat{i}+8\hat{j}+8\hat{k})\times10^{-3}$ N

📘 Vector Coulomb's Law F = kq₁q₂r⃗₁₂/|r⃗₁₂|³

The force on $q_2$ due to $q_1$ is along $\vec{r}_{12}$ (from $q_1$ to $q_2$). Since $q_1q_2<0$ (opposite charges), force is attractive. The negative product with negative direction vector gives positive result.

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Important Concepts

Displacement Vectorr⃗₁₂=position(q₂)−position(q₁)=(−1,−2,−2). |r⃗₁₂|=3m.

Vector FormF⃗=kq₁q₂r⃗₁₂/|r⃗₁₂|³. With k×q₁×q₂/|r|³=9e9×3e-6×(−4e-6)/27=−4×10⁻³.

ResultF⃗=−4×10⁻³×(−1,−2,−2)=(4,8,8)×10⁻³ N = option B.

Attractive Force CheckOpposite charges attract. Force on q₂ is toward q₁ = direction (1,2,2)/3. F=12×10⁻³×(1,2,2)/3=(4,8,8)×10⁻³ N ✓

FAQs

How to find r⃗₁₂?

⌄

r⃗₁₂=r₂−r₁=(1−2,1−3,1−3)=(−1,−2,−2). Points from q₁ toward q₂.

Why does force come out positive (toward q₁)?

⌄

kq₁q₂=9e9×3e-6×(−4e-6)=−108×10⁻³<0. Combined with negative r⃗₁₂=(−1,−2,−2): product is positive, pointing from q₂ toward q₁.

|r|=3?

⌄

√(1+4+4)=√9=3 m.

Magnitude check

⌄

F=k|q₁||q₂|/r²=9×10⁹×12×10⁻¹²/9=12×10⁻³ N. (4,8,8)×10⁻³: magnitude=√(16+64+64)×10⁻³=√144×10⁻³=12×10⁻³ ✓

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026.

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