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Ch 23: Electric Potential
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 23, Problem 17a

Point charges q1=+2.00q_1 = +2.00 μμC and q2=2.00q_2 = -2.00 μμC are placed at adjacent corners of a square for which the length of each side is 3.003.00 cm. Point aa is at the center of the square, and point bb is at the empty corner closest to q2q_2q2q_2. Take the electric potential to be zero at a distance far from both charges. (a) What is the electric potential at point a due to q1q_1 and q2q_2?

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1
Understand that the electric potential due to a point charge is given by the formula: V=kqr, where k is Coulomb's constant (k=8.99×10^9N·m2/C2), q is the charge, and r is the distance from the charge to the point where the potential is being calculated.
Calculate the distance from each charge to point a, which is the center of the square. Since the charges are at adjacent corners, the distance to the center can be found using the Pythagorean theorem: r=2×3.002 cm.
Convert the distance from centimeters to meters for consistency with SI units: r=2×0.032 m.
Calculate the electric potential at point a due to each charge separately using the formula: V=kqr. For q=2.00 μC, convert to Coulombs: q=2.00×10^-6 C.
Add the potentials due to each charge to find the total electric potential at point a. Since one charge is positive and the other is negative, their potentials will have opposite signs, and you will sum them algebraically.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential

Electric potential is the work done per unit charge in bringing a positive test charge from infinity to a point in space, without acceleration. It is a scalar quantity measured in volts and is crucial for understanding the energy landscape created by electric charges. The potential due to a point charge is given by V = kQ/r, where k is Coulomb's constant, Q is the charge, and r is the distance from the charge.
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Superposition Principle

The superposition principle states that the total electric potential at a point due to multiple charges is the algebraic sum of the potentials due to each charge individually. This principle is essential for calculating the net potential at a point when multiple charges are present, as it allows us to consider each charge's effect independently and then combine them.
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Geometry of the System

Understanding the geometry of the system is crucial for calculating distances between charges and points of interest. In this problem, the charges are placed at adjacent corners of a square, and the point of interest is at the center. The geometry helps determine the distances needed to calculate the electric potential using the formula V = kQ/r, where r is the distance from each charge to the point.
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Related Practice
Textbook Question

A particle with charge +4.20+4.20 nC is in a uniform electric field EE directed to the left. The charge is released from rest and moves to the left; after it has moved 6.006.00 cm, its kinetic energy is +2.20x106+2.20x10^{-6} J. What is the work done by the electric force?

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Textbook Question

Two stationary point charges +3.00+3.00 nC and +2.00+2.00 nC are separated by a distance of 50.050.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.010.0 cm from the +3.00+3.00-nC charge?

Textbook Question

Two point charges q1=+2.40q_1 = +2.40 nC and q2=6.50q_2 = -6.50 nC are 0.1000.100 m apart. Point AA is midway between them; point BB is 0.0800.080 m from q1q_1 and 0.0600.060 m from q2q_2 (Fig. E23.1923.19). Take the electric potential to be zero at infinity. Find the potential at point BB.

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Textbook Question

A small particle has charge 5.00-5.00 μμC and mass 2.00×1042.00\(\times\)10^{-4} kg. It moves from point AA, where the electric potential is VA=+200V_A = +200 V, to point BB, where the electric potential is VB=+800V_B = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.005.00 m/s at point AA. What is its speed at point BB? Is it moving faster or slower at BB than at AA? Explain.

Textbook Question

Two point charges of equal magnitude QQ are held a distance dd apart. Consider only points on the line passing through both charges. If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?).

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Textbook Question

Two point charges q1=+2.40q_1 = +2.40 nC and q2=6.50q_2 = -6.50 nC are 0.1000.100 m apart. Point AA is midway between them; point BB is 0.0800.080 m from q1q_1 and 0.0600.060 m from q2q_2 (Fig. E23.1923.19). Take the electric potential to be zero at infinity. Find the potential at point AA.

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