Two point charges … and … are placed at…

Single correct+4 / -1Answer matches NTA’s final key

Two point charges $q_1 = 3\,\mu C$ and $q_2 = -4\,\mu C$ are placed at points $\left(2\hat i + 3\hat j + 3\hat k\right)$ and $\left(\hat i + \hat j + \hat k\right)$ respectively. Force on charge $q_2$ is ______ N. $\left(\text{Take } \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI Units}\right)$

$\left(12\hat i + 24\hat j + 24\hat k\right) \times 10^{-3}$
$\left(4\hat i + 8\hat j + 8\hat k\right) \times 10^{-3}$
$\left(3\hat i + 6\hat j + 6\hat k\right) \times 10^{-3}$
$\left(-4\hat i - 8\hat j - 8\hat k\right) \times 10^{-3}$

Solution

The vector from $q_1$ to $q_2$ is $\vec r = (1, 1, 1) - (2, 3, 3) = (-1, -2, -2)$ , with $|\vec r| = 3$ m.

$\vec F_2 = \frac{kq_1q_2}{r^3}\vec r = \frac{9 \times 10^9 \times 3 \times 10^{-6} \times (-4 \times 10^{-6})}{27}(-1, -2, -2) = 4 \times 10^{-3}(1, 2, 2).$

The charges attract, so the force on $q_2$ points towards $q_1$ , which is what the sign shows.

Answer: $\left(4\hat i + 8\hat j + 8\hat k\right) \times 10^{-3}$ N.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Electrostatics

Solution

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