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Chapter 25: Q. 39 (page 711)

A +3.0 nC charge is at x = 0 cm and a -1.0 nC charge is at x = 4 cm. At what point or points on the x-axis is the electric potential zero?

Short Answer

Expert verified

The points are 3 cm and 6 cm

Step by step solution

01

Definition of electric potential

Assume for $q_{1}$ , x=0 and for $q_{2}$ , x=+4.0 cm

When $x < 0$ positive charges> negative charge. due to this reason, they can not be cancelled.

At when $0 < x < 4$

localid="1648130310708" $V = 1 / (4 π ε_{0}) (\frac{q_{1}}{r_{0}} + \frac{q_{2}}{r_{4}}) q_{1} = 3 \times 10^{- 9} C q_{2} = 1 \times 10^{- 9} C r_{0} = x r_{4} = 4 - x V = 1 / (4 π ε_{0}) (\frac{3 \times 10^{- 9} C}{x} + \frac{1 \times 10^{- 9} C}{4 - x}) A t V = 0 x = 3.0 c m$

[Note: the value of electric potential for two charges is 0, V=0 V]

02

Definition of electric potential

At when x> 4

$\begin{array}{rcl} V & = & 1 / (4 π ε_{0}) (\frac{q_{1}}{r_{0}} + \frac{q_{2}}{r_{4}}) \\ q_{1} & = & 3 \times 10^{- 9} C \\ q_{2} & = & 1 \times 10^{- 9} C \\ r_{0} & = & x \\ r_{4} & = & x - 4 \\ V & = & 1 / (4 π ε_{0}) (\frac{3 \times 10^{- 9} C}{x} + \frac{1 \times 10^{- 9} C}{x - 4}) \\ A t V & = & 0 \\ x & = & 6.0 c m \end{array}$

[Note: the value of electric potential for two charges is 0, V=0 V]

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Most popular questions from this chapter

In Problems 74 through 76 you are given the equation(s) used to solve a problem. For each of these,

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem.

75. $\frac{1}{2} (1.67 \times 10^{- 27} k g) {(2.5 \times 10^{6} m / s)}^{2} + 0 = \frac{1}{2} (1.67 \times 10^{- 27} k g) {v_{i}}^{2} + \frac{(9.0 \times 10^{9} N m^{2} / C^{2}) (2.0 \times 10^{- 9} C) (1.60 \times 10^{- 19} C)}{0.0010 m}$

The electric field strength is $20, 000 N / C$ inside a parallel-plate capacitor with a $1.0 m m$ spacing. An electron is released from rest at the negative plate. What is the electron’s speed when it reaches the positive plate?

One form of nuclear radiation, beta decay, occurs when a neutron changes into a proton, an electron, and a neutral particle called a neutrino: $n \to p^{+} + e^{-} + v$ where $n$ is the symbol for a neutrino. When this change happens to a neutron within the nucleus of an atom, the proton remains behind in the nucleus while the electron and neutrino are ejected from the nucleus. The ejected electron is called a beta particle. One nucleus that exhibits beta decay is the isotope of hydrogen $3 H$ , called tritium, whose nucleus consists of one proton (making it hydrogen) and two neutrons (giving tritium an atomic mass $m = 3 u$ ). Tritium is radioactive, and it decays to helium: ${}^{3}H \to {}^{3}H e + e^{-} + n$

a. Is charge conserved in the beta decay process? Explain.

b. Why is the final product a helium atom? Explain.

c. The nuclei of both ${}^{3}H and {}^{3}H e$ have radii of $1.5 \times 10^{- 15} m$ . With what minimum speed must the electron be ejected if it is to escape from the nucleus and not fall back?

Two 2.0 g plastic buttons each with +50 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to +250 nC.

All three are released simultaneously.

a. How many interactions are there that have a potential energy?

b. What is the final speed of each button?

In Problems 74 through 76 you are given the equation(s) used to solve a problem. For each of these,

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem.

74. $\frac{(9.0 \times 10^{9} N m^{2} / C^{2}) q_{1} q_{2}}{0.030 m} = 90 \times 10^{- 6} J q_{1} + q_{2} = 40 n C$

See all solutions

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