Question

Two positive charges, each equal to \(Q,\) are placed a distance \(2 d\) apart. A third charge, \(-0.2 Q\), is placed exactly halfway between the two positive charges and is displaced a distance \(x \ll d\) (that is, \(x\) is much smaller than \(d\) ) perpendicular to the line connecting the positive charges. What is the force on this charge? For \(x \ll d\), how can you approximate the motion of the negative charge?

Short answer

In this problem, we find the force acting on a negative charge \(-0.2Q\) placed between two positive charges \(Q\) in an unequal triangle configuration. Using Coulomb's Law and the superposition principle, we calculated the individual forces acting on the negative charge due to the positive charges and obtained the net force to be: \(F_{3, net} = 0.4 \dfrac{kQ^2}{d^2} (1 - 0.5 (\frac{x}{d})^2)\) Since \(x \ll d\), the force is approximately constant, and the motion of the charge can be considered as simple harmonic motion.

Step-by-step solution

Calculate the forces between individual charges

Let's label the charges as follows: Positive charges, \(Q_1\) and \(Q_2\), and the negative charge, \(Q_3\). Using Coulomb's Law, we can find the magnitude of the force acting between each pair of these charges: \(F_{ij} = \dfrac{k |Q_iQ_j|}{r_{ij}^2}\), where \(F_{ij}\) is the force between the charges \(Q_i\) and \(Q_j\), \(r_{ij}\) is the distance between them, and \(k\) is the electrostatic constant.

Calculate individual forces on the negative charge

We need to find the force acting on \(Q_3\). To do this, we can calculate the individual forces acting on it due to \(Q_1\) and \(Q_2\), and then find the vector sum of those forces. Since \(Q_3\) is exactly halfway between \(Q_1\) and \(Q_2\), the distance between \(Q_3\) and each positive charge is \(d\). Let's calculate each force using the formula from step 1: \(F_{13} = \dfrac{k |Q_1Q_3|}{r_{13}^2} = \dfrac{k|Q(-0.2Q)|}{d^2} = 0.2 \dfrac{k Q^2}{d^2}\) \(F_{23} = \dfrac{k |Q_2Q_3|}{r_{23}^2} = \dfrac{k|Q(-0.2Q)|}{d^2} = 0.2 \dfrac{kQ^2}{d^2}\) Here, we find that both forces have the same magnitude.

Find the net force on the negative charge

Now that we have found the magnitude of the individual forces, we need to find the vector sum of them. Due to symmetry, the component of the net force in the direction of the line connecting \(Q_1\) and \(Q_2\) cancels out. The remaining component of the net force is perpendicular to that line. The net force can be calculated as: \(F_{3, net} = 2 * F_{13} \cos{\theta}\), where \(\theta\) is the angle between the direction of \(F_{13}\) and the perpendicular direction. Using the small-angle approximation for the cosine (\(\cos \theta \approx 1 - \frac{1}{2}\theta^2\)), we can simplify the equation.

Simplify the net force with small-angle approximation

To analyze the motion, we need to assume the displacement \(x\) much smaller than \(d\), or \(x \ll d\). Since we are working in the limit of small angles, we'll make use of the small-angle approximation. We can form the small angle \(\theta\) using the displacement \(x\) and the distance \(d\) along with the right triangle, \(\tan{\theta} \approx \theta = \dfrac{x}{d}\). Now, we can substitute this expression into the small-angle approximation: \(\cos{\theta} \approx 1 - \frac{1}{2}\theta^2 = 1 - 0.5 (\frac{x}{d})^2\). Substitute this approximation into the net force expression from Step 3: \(F_{3, net} = 2 * 0.2 \dfrac{kQ^2}{d^2} (1 - 0.5 (\frac{x}{d})^2)\)

Simplify and obtain the final expression for force

After simplifying the net force on \(Q_3\): \(F_{3, net} = 0.4 \dfrac{kQ^2}{d^2} (1 - 0.5 (\frac{x}{d})^2)\) This is the force acting on the negative charge \(-0.2Q\). The negative charge experiences a force away from the line connecting the two positive charges, with the magnitude given by this expression. Since \(x \ll d\), the force is approximately constant and the motion of the charge can be considered to involve simple harmonic motion.