To solve the problem, we need to determine the correct position of a third charge so that it experiences no net force due to the two given charges. We will use the concept of electrostatic forces and the principle of superposition.
The two charges are: \(q_1 = 5 \times 10^{-11} \mathrm{C}\) and \(q_2 = -2.7 \times 10^{-11} \mathrm{C}\). The distance between them is \(d_{12} = 0.2 \ \mathrm{m}\).
Let the third charge \(q_3\) be placed at a distance \(x\) from the second charge \(q_2\), such that \(q_3\) experiences no net force. For this, the magnitudes of the forces due to \(q_1\) and \(q_2\) on \(q_3\) must be equal.
The force exerted by \(q_1\) on \(q_3\) is given by Coulomb's law: \(F_{13} = k \frac{|q_1 \cdot q_3|}{(d_{12} + x)^2}\).
The force exerted by \(q_2\) on \(q_3\) is: \(F_{23} = k \frac{|q_2 \cdot q_3|}{x^2}\).
According to the condition of no net force, these two forces must be equal: \(F_{13} = F_{23}\).
Setting them equal, we have: \(k \frac{|q_1 \cdot q_3|}{(d_{12} + x)^2} = k \frac{|q_2 \cdot q_3|}{x^2}\).
Cancel the common terms and simplify:
\(\frac{|q_1|}{(d_{12} + x)^2} = \frac{|q_2|}{x^2}\)
Plug in the values \(|q_1| = 5 \times 10^{-11} \mathrm{C}\) and \(|q_2| = 2.7 \times 10^{-11} \mathrm{C}\):
\(\frac{5 \times 10^{-11}}{(0.2 + x)^2} = \frac{2.7 \times 10^{-11}}{x^2}\)
Solve for \(x\):
Cross-multiply and simplify:
\(5x^2 = 2.7(0.2 + x)^2\)
Expand the right-hand side:
\(5x^2 = 2.7(0.04 + 0.4x + x^2)\)
Distribute:
\(5x^2 = 0.108 + 1.08x + 2.7x^2\)
Rearrange and simplify:
\(2.3x^2 - 1.08x - 0.108 = 0\)
Use the quadratic formula to solve for \(x\):
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Where \(a = 2.3\), \(b = -1.08\), and \(c = -0.108\).
Calculate:
\(x = \frac{-(-1.08) \pm \sqrt{(-1.08)^2 - 4 \cdot 2.3 \cdot (-0.108)}}{2 \cdot 2.3}\) \(x = \frac{1.08 \pm \sqrt{1.1664 + 0.9936}}{4.6}\) \(x = \frac{1.08 \pm \sqrt{2.16}}{4.6}\) \(x \approx \frac{1.08 \pm 1.47}{4.6}\)
The positive solution gives:
\(x \approx \frac{2.55}{4.6} \approx 0.556 \mathrm{m}\)
Therefore, the distance from the second charge at which the third charge should be placed is 0.556 m.
A 10 $\mu\text{C}$ charge is placed in an electric field of $ 5 \times 10^3 \text{N/C} $. What is the force experienced by the charge?