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

Short Answer

Expert verified
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 is +2.00 nC, (b) 45.0 N/C in the -x -direction is -2.00 nC.

Step by step solution

01

Calculate the electric field due to the fixed charge

At the origin, the electric field due to the fixed charge at x = 1.20 m, E1 = k|q1|/r鈧伮, where q1 = -3.00 nC, r鈧 = 1.20 m. Substituting the values, E1 = (9 x 10^9 N.m虏/C虏 * 3 x 10^-9 C) / (1.2 m)虏 = -15 N/C in the -x direction, as field due to a negative charge is towards the charge.
02

Calculate the electric field due to the second charge for the resultant field to be in +x direction

For the resultant field at the origin to be 45.0 N/C in the +x direction, the electric field due to the second charge, E2 = E - E1, where E is the resultant field. Substituting the values, E2 = 45.0 N/C - (-15 N/C) = 60 N/C in the +x direction. Now, using the formula for electric field, E = k|q|/r虏, the charge Q = E2.r鈧偮/k. Substituting the values, Q = (60 N/C * (0.6 m)虏) / (9 x 10^9 N.m虏/C虏) = 2.00 nC. Since E2 is in the +x direction, Q must be positive.
03

Calculate the electric field due to the second charge for the resultant field to be in -x direction

For the resultant field at the origin to be 45.0 N/C in the -x direction, the electric field due to the second charge, E2 = E1 - E, where E is the resultant field. Substituting the values, E2 = (-15 N/C) - 45.0 N/C = -60 N/C in the -x direction. Now, using the formula for electric field, E = k|q|/r虏, the charge Q = E2.r鈧偮/k. Substituting the values, Q = (-60 N/C * (0.6 m)虏) / (9 x 10^9 N.m虏/C虏) = -2.00 nC. Since E2 is in the -x direction, Q must be negative.

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Key Concepts

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

Coulomb's Law
Coulomb's law is the cornerstone of electrostatics, illustrating how electric charges interact with each other. It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is given by
\( F = k \frac{|q_1q_2|}{r^2} \),
where \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance separating them, and \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \) N.m虏/C虏.
Coulomb's law applies to point charges - idealized charges located at a single point in space. Despite this simplification, Coulomb's law provides a fundamental understanding of how charges exert forces on each other and is the basis for calculating electric fields, as seen in the exercise provided.
Electric Force
The electric force is the force experienced by a charge in the presence of another charge or an electric field. When charges are stationary, this is referred to as the electrostatic force, as per Coulomb's law. The direction of the force depends on the nature of the charges; like charges repel each other, while opposite charges attract.

Application in Exercise

In the exercise scenario, the force is determined by the electric field created by the charges. The electric field, denoted as \( E \), at a point in space is defined as the force per unit charge experienced by a small positive test charge placed at that point. It has both magnitude and direction, and for a point charge, it is calculated by rearranging Coulomb鈥檚 law to
\( E = k \frac{|q|}{r^2} \).
In the context of the exercise, the electric field due to the point charge is used to determine the necessary characteristics of a second charge in order to achieve a specific resultant electric field at the origin.
Superposition Principle
The superposition principle is a fundamental concept in physics, particularly in the realm of electrostatics. It posits that when multiple charges are present, the total electric force at a specific point is the vector sum of the electric forces produced by each individual charge acting independently. This principle allows us to calculate complex fields by breaking them down into simpler components.

Application in Exercise

The principle is applied in the exercise by adding the electric fields due to individual charges when evaluating the resultant electric field at the origin. Each charge contributes to the total field, and their individual fields are algebraically added, considering both magnitude and direction, to find the combined effect. Thus, the second charge in the exercise must be chosen in such a way that it creates an electric field that, when combined with the field from the initial charge, results in the desired net electric field at the origin.

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Most popular questions from this chapter

(a) What must the charge (sign and magnitude) of a \(1.45 \mathrm{~g}\) particle be for it to remain stationary when placed in a downwarddirected electric field of magnitude \(650 \mathrm{~N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of \(2.50 \mathrm{~mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

A point charge is placed at each corner of a square with side length a. All charges have magnitude \(q\). Two of the charges are positive and two are negative (Fig. E21.38). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of \(q\) and \(a\) ?

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Ammonia molecules in the gas phase are placed in a uniform electric field \(\vec{E}\) with magnitude \(1.6 \times 10^{6} \mathrm{~N} / \mathrm{C}\). (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\vec{E}\) from parallel to perpendicular? (b) At what absolute temperature \(T\) is the average translational kinetic energy \(\frac{3}{2} k T\) of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

A uniform line of charge with length \(20.0 \mathrm{~cm}\) is along the \(x\) -axis, with its midpoint at \(x=0 .\) Its charge per length is \(+4.80 \mathrm{nC} / \mathrm{m}\) A small sphere with charge \(-2.00 \mu \mathrm{C}\) is located at \(x=0, y=5.00 \mathrm{~cm}\) What are the magnitude and direction of the force that the charged sphere exerts on the line of charge?

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