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Three point charges are arranged on a line. Charge \(q_3 = +\)5.00 nC and is at the origin. Charge \(q_2 = -\)3.00 nC and is at \(x = +\)4.00 cm. Charge \(q_1\) is at \(x = +\)2.00 cm. What is \(q_1\) (magnitude and sign) if the net force on \(q_3\) is zero?

Short Answer

Expert verified
Charge \( q_1 = +0.75 \) nC to balance forces on \( q_3 \).

Step by step solution

01

Identifying Forces on q_3

First, let's identify the forces acting on charge \( q_3 \). There are two charges interacting with \( q_3 \): \( q_1 \) located at \( x = +2.00 \) cm and \( q_2 \) located at \( x = +4.00 \) cm. Each will exert a force on \( q_3 \) based on Coulomb's law.
02

Applying Coulomb's Law

Use Coulomb's law to calculate the force each charge exerts on \( q_3 \). Coulomb's Law is given by \( F = k \frac{|q_1 q_3|}{r^2} \), where \( k \) is the electrostatic constant \( 8.99 \times 10^9 \) N·m²/C², \( q_1 \) and \( q_3 \) are the charges, and \( r \) is the distance between them. Apply this law for forces \( F_{13} \) and \( F_{23} \) exerted by \( q_1 \) and \( q_2 \), respectively.
03

Setting Net Force to Zero

For the net force on \( q_3 \) to be zero, the force by \( q_1 \) on \( q_3 \) must be equal and opposite to the force by \( q_2 \) on \( q_3 \). Mathematically, this is \( F_{13} + F_{23} = 0 \). Express this in terms of the known quantities and solve for \( q_1 \).
04

Finding the Distance and Direction

Calculate the distances: \( r_{13} \) (between \( q_1 \) and \( q_3 \)) is \( 2.00 \) cm, and \( r_{23} \) (between \( q_2 \) and \( q_3 \)) is \( 4.00 \) cm. Also, note the direction: \( q_3 \) is positively charged, so it will be attracted towards \( q_2 \) (negative charge) and repelled by positive charges.
05

Solve for q_1

Use the modified equation: \[ k \frac{|q_1 \, q_3|}{(0.02)^2} = k \frac{|q_2 \, q_3|}{(0.04)^2} \]Cancel \( k \) and \( |q_3| \) as they appear on both sides, yielding:\[ \frac{|q_1|}{0.0004} = \frac{|-3.00 \, \text{nC}|}{0.0016} \] Solve for \( |q_1| \):\[ |q_1| = -3.00 \, \text{nC} \times \frac{0.0004}{0.0016} = -0.75 \, \text{nC} \]Charge \( q_1 \) needs to be positive to produce a repulsive force that counterbalances the attraction due to \( q_2 \). Hence, \( q_1 = +0.75 \, \text{nC} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
Electric force is the key player in the interaction between charged particles. This force can be attractive or repulsive depending on the nature of the charges involved. According to Coulomb's Law, the electric force \( F \) between two point charges is calculated by the formula:
  • \( F = k \frac{|q_1 \cdot q_2|}{r^2} \)
where:
  • \( k \) is the electrostatic constant, approximately \( 8.99 \times 10^9 \) N\cdot m²/C².
  • \( q_1 \) and \( q_2 \) are the amounts of the two point charges.
  • \( r \) is the distance between the charges.
The direction of the electric force is determined by the signs of the charges:
  • If both charges are of the same sign, the force is repulsive, pushing them apart.
  • If the charges are of opposite signs, the force is attractive, pulling them together.
In the case of our exercise, the electric forces between the different charges must balance out for the net force on charge \( q_3 \) to be zero. This concept is crucial in figuring out the appropriate magnitude and sign of any unknown charge.
Electrostatics
Electrostatics deals with the study of stationary or slow-moving electric charges. It is an essential area of physics providing the basis for understanding electric interactions at rest. Electrostatic phenomena arise from the forces that electric charges exert on one another. A fundamental principle in electrostatics is that a charge can affect other charges in its vicinity via electric fields, without needing a medium to travel through.
In our given problem, we analyze the interaction of these charges using the principles of electrostatics. These interactions focus on understanding the balance between forces exerted by each point charge on one another when they're located at fixed positions in space. The objective here was to achieve static equilibrium, where the net force experienced by charge \( q_3 \) sums to zero meaning:
  • The attractive force from \( q_2 \) (negative charge) is counterbalanced by the repulsive force from \( q_1 \) (positive charge).
  • Thus, applying electrostatic principles can determine characteristics like charge magnitude and sign.
Point Charges
Point charges are idealizations used to simplify and analyze problems involving electrostatic forces. A point charge is assumed to be a charged particle with negligible size compared to distances involved in the problem, allowing for the concentration of an entire charge at a single point. This simplification helps in calculating the force and potential energy between particles.
In the context of the exercise:
  • \( q_1 \), \( q_2 \), and \( q_3 \) are treated as point charges along a line. This enables the use of Coulomb's law easily, based only on their distances and amounts of charge.
  • The exercise involves calculating these interactions at constant distances of 2 cm and 4 cm to ensure correct force balancing.
Dealing with point charges means realizing that these charges can exert significant forces even though their dimensions are negligibly small. They make calculations easier by reducing complex charge distributions to simple, manageable forms.

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Most popular questions from this chapter

A charge \(q_1 = +\)5.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge \(q_2 = -\)2.00 nC is placed on the positive \(x\)-axis at \(x = \)4.00 cm. (a) If a third charge \(q_3 = +\)6.00 nC is now placed at the point \(x =\) 4.00 cm, \(y =\) 3.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

A \(-\)3.00-nC point charge is on the \(x\)-axis at \(x =\) 1.20 m. A second point charge, \(Q,\) is on the \(x\)-axis at -0.600 m. What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 N\(/\)C in the \(+\)x-direction, (b) 45.0 N\(/\)C in the \(-\)x-direction?

Two small plastic spheres are given positive electric charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Two small spheres, each carrying a net positive charge, are separated by \(0.400 m\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (\(charge \space q_1\)) at the origin and the other sphere (\(charge \space q_2\)) at \(x = +\)0.400 m. Available to you are a third sphere with net charge \(q_3 = 4.00 \times 10^{-6}\) C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\)-axis at \(x =\) 0.200 m; you measure the net force on it to be 4.50 N in the \(+ x\)-direction. Then you move the third sphere to \(x = +\)0.600 m and measure the net force on it now to be 3.50 N in the \(+ x\)-direction. (a) Calculate \(q_1\) and \(q_2\). (b) What is the net force (magnitude and direction) on \(q_3\) if it is placed on the \(x\)-axis at \(x = -\)0.200 m? (c) At what value of \(x\) (other than \(x = \pm \infty\)) could \(q_3\) be placed so that the net force on it is zero?

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

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