Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

The formula to calculate the electrical potential 'V' produced by a single point charge 'Q' at a distance 'r' is: \[V = \frac{kQ}{r}\] Where 'k' is the Coulomb's constant: \(k = 8.9875 × 10^9 Nm^2C^{-2}\). Now, let's find the potential at a point 'x' where the electrical potential will be zero.

The net potential at a point x due to the two point charges is the algebraic sum of the potential due to each charge: \(V(x) = V_1(x) + V_2(x)\) The potential due to charge \(q_1 = 2 \mu C\) at point x is: \(V_1(x) = \frac{kq_1}{x}\) and the potential due to charge \(q_2 = -1 \mu C\) at point x is: \(V_2(x) = \frac{kq_2}{(6-x)}\) Since we are looking for the points where the electrical potential is zero, we set the net potential equal to zero: \(V(x) = 0\)

Using the formulas from step 2, equate \(V_1(x) + V_2(x) = 0\): \[\frac{kq_1}{x} + \frac{kq_2}{(6-x)} = 0\] Now, substitute the values \(q_1 = 2 \mu C\) and \(q_2 = -1 \mu C\): \[\frac{2k}{x} - \frac{k}{(6-x)} = 0\] Clear the denominators by multiplying through by x(6-x) and simplify: \[2k(6 - x) - kx = 0\] \[k(12 - 3x) = 0\] Since the constant 'k' cannot be zero, the remaining term must be zero: \[12 - 3x = 0\] Solve for x: \[x = 4\] So, due to the two charges, the electrical potential will be zero at point x = 4. This means that the correct answer is (C) x = 4. Note that there may be other points where the electrical potential is zero. However, those points would be hidden in the other choices (A), (B), and (D). Therefore, the correct choice is (C), where x = 4.