Two point charges q1 = 3 μC and q2 = -4 μC are placed at points (2i + 3j + 3k) and (i + j + k) respectively. Force on charge q2 is _______ N. (Take 1/(4πε0) = 9 x 10^9 SI Units)

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q41 – JEE-MAIN 2026

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Home Physics Electric Charges and FieldsQuestion #101

Two point charges

q_{1} = 3 μ C

and

q_{2} = - 4 μ C

are placed at points

(2 \hat{i} + 3 \hat{j} + 3 \hat{k})

and

(\hat{i} + \hat{j} + \hat{k})

respectively. Force on charge

q_{2}

is _______ N. (Take

\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9}

SI Units)

Clue 1 of 3: Coulomb's Law and Vector Nature

Remember that electrostatic force is a vector quantity, and its direction depends on the signs of the charges and the relative positions of the charges.

🧠 Full Solution Path

Step 1: Determine the displacement vector and its magnitude✦ Active

The position vector of charge $q_{1}$ is $\vec{r_{1}} = 2 \hat{i} + 3 \hat{j} + 3 \hat{k}$ and for $q_{2}$ is $\vec{r_{2}} = \hat{i} + \hat{j} + \hat{k}$ . The displacement vector from $q_{1}$ to $q_{2}$ is ${\vec{r}}_{12} = \vec{r_{2}} - \vec{r_{1}}$ .

{\vec{r}}_{12} = (\hat{i} + \hat{j} + \hat{k}) - (2 \hat{i} + 3 \hat{j} + 3 \hat{k}) = - \hat{i} - 2 \hat{j} - 2 \hat{k}

The magnitude of this displacement vector is:

| {\vec{r}}_{12} | = \sqrt{(- 1)^{2} + (- 2)^{2} + (- 2)^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 m

The unit vector in this direction is:

{\hat{r}}_{12} = \frac{{\vec{r}}_{12}}{| {\vec{r}}_{12} |} = \frac{- \hat{i} - 2 \hat{j} - 2 \hat{k}}{3}

Step 2: Apply Coulomb's Law in vector form○ Expand

The force on charge $q_{2}$ due to $q_{1}$ is given by Coulomb's Law:

{\vec{F}}_{21} = k \frac{q_{1} q_{2}}{| {\vec{r}}_{12} |^{2}} {\hat{r}}_{12}

Given $q_{1} = 3 μ C = 3 \times 10^{- 6} C$ , $q_{2} = - 4 μ C = - 4 \times 10^{- 6} C$ , and $k = 9 \times 10^{9} N m^{2} / C^{2}$ . Substitute these values:

{\vec{F}}_{21} = (9 \times 10^{9}) \frac{(3 \times 10^{- 6}) (- 4 \times 10^{- 6})}{(3)^{2}} (\frac{- \hat{i} - 2 \hat{j} - 2 \hat{k}}{3})

{\vec{F}}_{21} = (9 \times 10^{9}) \frac{- 12 \times 10^{- 12}}{9} (\frac{- \hat{i} - 2 \hat{j} - 2 \hat{k}}{3})

{\vec{F}}_{21} = (- 12 \times 10^{- 3}) (\frac{- \hat{i} - 2 \hat{j} - 2 \hat{k}}{3})

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

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

💡 Teacher's Secret Hint

Pay close attention to the signs of the charges and the direction of the displacement vector. A negative product of charges indicates an attractive force, meaning the force vector will be opposite to the displacement vector from $q_{1}$ to $q_{2}$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 41 – JEE-MAIN 2026

Clue 1 of 3: Coulomb's Law and Vector Nature

🧠 Full Solution Path