Two point charges of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)

The electric field produced by a point charge is given by the formula \(E=k\frac{q}{r^2}\), where \(E\) is the electric field, \(k\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance from the charge to the point where the electric field is calculated. In this exercise, we have two point charges: \(q_1=+16c\) and \(q_2=-9c\), so the electric field produced by each charge will be given by \(E_1 = k\frac{q_1}{r_1^2}\), and \(E_2=k\frac{q_2}{r_2^2}\).

By the principle of superposition, the resultant electric field at any point is given by the vector sum of the electric fields produced by each point charge. In this problem, we want to find a point where the resultant electric field is zero, meaning the electric fields produced by the charges must cancel each other. Without loss of generality, place the +16c charge at the origin and the -9c charge on the positive x-axis at a distance of 8cm. Then, we need a point that is equidistant from both charges along the x-axis so that the electric fields produced by both charges will be equal in magnitude and opposite in direction. Since the -9c charge is greater in magnitude, the location where the electric field cancels out is between the two charges.

The electric fields produced by the two charges at their respective distances should be equal in magnitude and opposite in direction, so we can set up the equation: \(k\frac{q_1}{r_1^2} = k\frac{-q_2}{r_2^2}\) where \(r_1\) is the distance from -9c charge to the point, and \(r_2\) is the distance from +16c charge to the point (or the distance between the two charges minus the distance from the -9c charge to the point).

Substitute the given values into the equation and solve for \(r_1\): \(\frac{16}{r_1^2} = \frac{9}{(8-r_1)^2}\) We can cross multiply and simplify the equation: \(16(8-r_1)^2 = 9r_1^2\) Let's expand the squares: \(16(64 - 16r_1 + r_1^2) = 9r_1^2\) Combine like terms: \(144r_1^2 - 256r_1 + 1024 = 0\) Now, we can use either the Quadratic Formula or factor the equation to find \(r_1\). In this case, factoring is faster: \(144(r_1 - 4)^2 = 0\) \(r_1 = 4\) However, since we placed the -9c charge on the positive x-axis at a distance of 8cm, the distance of a point from the -9c charge is between it and the +16c charge: \(r_1 = 8 - 4 = 4 \mathrm{~cm}\multirow{2}{*}{ }\) Thus, the distance from the -9c charge at which the resultant electric field is zero is 4cm, which is not found in the given options (A: 24cm, B: 9cm, C: 16cm, and D: 35cm). It is possible that there is an error with the options provided or the question needs revision.