Q.41. Two point charges q₁ = 3µC and q₂=-4µC are placed at points (2+3+3) and (++) respectively. Force on charge q₂ is N. (Take = 9×10⁹ SI Units)

Sol. $$\vec{r}$$ = $$\vec{r_2}$$ - $$\vec{r_1}$$ = ($$\hat{i}$$+$$\hat{j}$$+$$\hat{k}$$) - (2$$\hat{i}$$+3$$\hat{j}$$+3$$\hat{k}$$) = -$$\hat{i}$$-2$$\hat{j}$$-2$$\hat{k}$$ Force = $$\frac{kq_1q_2}{r^3}$$ $$\vec{r}$$ = $$\frac{9 \times 10^9 \times (3) \times (-4) \times 10^{-12}}{27}$$ ($$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$) ⇒ 4×10⁻³[$$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$]

(12$$\hat{i}$$+24$$\hat{j}$$+24$$\hat{k}$$)×10⁻³

(4$$\hat{i}$$+8$$\hat{j}$$+8$$\hat{k}$$)×10⁻³

(3$$\hat{i}$$+6$$\hat{j}$$+6$$\hat{k}$$)×10⁻³

(-4$$\hat{i}$$-8$$\hat{j}$$-8$$\hat{k}$$)×10⁻³

What is the correct answer for JEE Main Physics Q41?

The correct answer is B. Sol. $$\vec{r}$$ = $$\vec{r_2}$$ - $$\vec{r_1}$$ = ($$\hat{i}$$+$$\hat{j}$$+$$\hat{k}$$) - (2$$\hat{i}$$+3$$\hat{j}$$+3$$\hat{k}$$) = -$$\hat{i}$$-2$$\hat{j}$$-2$$\hat{k}$$ Force = $$\frac{kq_1q_2}{r^3}$$ $$\vec{r}$$ = $$\frac{9 \times 10^9 \times (3) \times (-4) \times 10^{-12}}{27}$$ ($$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$) ⇒ 4×10⁻³[$$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$]

Q.41. Two point charges q₁ = 3µC and q₂=-4µC are placed at points (2+3+3) and (++) respectively. Force on charge q₂ is N. (Take = 9×10⁹ SI Units)

The correct answer is B. Sol. $$\vec{r}$$ = $$\vec{r_2}$$ - $$\vec{r_1}$$ = ($$\hat{i}$$+$$\hat{j}$$+$$\hat{k}$$) - (2$$\hat{i}$$+3$$\hat{j}$$+3$$\hat{k}$$) = -$$\hat{i}$$-2$$\hat{j}$$-2$$\hat{k}$$ Force = $$\frac{kq_1q_2}{r^3}$$ $$\vec{r}$$ = $$\frac{9 \times 10^9 \times (3) \times (-4) \times 10^{-12}}{27}$$ ($$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$) ⇒ 4×10⁻³[$$\hat{i}$$+2$$\hat{j}$$+2$$\hat{k}$$]

What are the options for this JEE Main Physics question?

A. (12$$\hat{i}$$+24$$\hat{j}$$+24$$\hat{k}$$)×10⁻³ · B. (4$$\hat{i}$$+8$$\hat{j}$$+8$$\hat{k}$$)×10⁻³ · C. (3$$\hat{i}$$+6$$\hat{j}$$+6$$\hat{k}$$)×10⁻³ · D. (-4$$\hat{i}$$-8$$\hat{j}$$-8$$\hat{k}$$)×10⁻³

JEE Main 2026 Physics PYQ · With Answer · Free

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Q.41. Two point charges q₁ = 3µC and q₂=-4µC are placed at points (2$$\hat{i}$$+3$$\hat{j}$$+3$$\hat{k}$$) and ($$\hat{i}$$+$$\hat{j}$$+$$\hat{k}$$) respectively. Force on charge q₂ is N. (Take $$\frac{1}{4\pi \epsilon_0}$$ = 9×10⁹ SI Units)

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