Question

A point charge \(q_1 = 4q_0\) is placed at origin. Another point charge \(q_2 = –q_0\) is placed at \(x = 12\text{ cm}\). Charge of proton is q₀. The proton is placed on x axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is _____ cm.

Hint:

For force equilibrium problems:
• Use Coulomb’s law to equate the forces due to charges on opposite sides.
• Solve for the distance r and adjust for the final position based on geometry.

Answer 1

Correct Answer: 24

To find the position where the electrostatic force on the proton is zero, we need to analyze the forces due to charges \(q_1\) and \(q_2\) on the proton placed at position \(x\) along the x-axis.
Given:
\(q_1 = 4q_0\) at \(x_1 = 0\)
\(q_2 = -q_0\) at \(x_2 = 12\text{ cm}\)

Assume the proton is located at \(x = X\) cm.
The force due to \(q_1\) at a distance \(X\) is given by:
\(F_1 = \frac{k \cdot |4q_0 \cdot q_0|}{X^2}\)
The force due to \(q_2\) at distance \(12-X\) is:
\(F_2 = \frac{k \cdot |(-q_0) \cdot q_0|}{(12-X)^2}\)
For the forces to balance out, \(F_1 = F_2\). Thus:
\(\frac{4kq_0^2}{X^2} = \frac{kq_0^2}{(12-X)^2}\)
Cancel \(kq_0^2\):
\(\frac{4}{X^2} = \frac{1}{(12-X)^2}\)
\(4(12-X)^2 = X^2\)
Expanding and simplifying:
\(4(144-24X+X^2) = X^2\)
\(576 - 96X + 4X^2 = X^2\)
\(3X^2 - 96X + 576 = 0\)
Divide the equation by 3:
\(X^2 - 32X + 192 = 0\)
Solve using the quadratic formula \(X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1, b=-32, c=192\):
\(X = \frac{32 \pm \sqrt{1024 - 768}}{2}\)
\(X = \frac{32 \pm \sqrt{256}}{2}\)
\(X = \frac{32 \pm 16}{2}\)
\(X = 24 \text{ or } 8\)
The proton cannot be positioned between \(q_1\) and \(q_2\) since a negative charge attracts the proton while a positive one repels, resulting in zero net force only outside the segment \( (0,12) \).
Therefore, the position that satisfies the conditions is \(X = 24\text{ cm}\), in the range [24, 24].