Chapter 25: Q. 36 (page 711)

A 5.0-cm-diameter metal ball has a surface charge density of $10 m C / m 2$ . How much work is required to remove one electron from this ball?

Short Answer

Expert verified

The work is needed to remove one electron from the ball is $- 4.5 \times 10^{- 15} J$

Step by step solution

Definition of potential charge

The equation of the potential of a charge is

$V = 1 / (4 π ε_{0}) q / r$

as the charge is sphere so the $q = 4 π r^{2} ρ$

Applying the data $V = \frac{r ρ}{ε_{0}}$

theory of work energy

Removal one electron from the ball, the work needs

As $W = e V$ $W = e V$

localid="1648302583389" $\begin{array}{rcl} ρ & = & 10 μ C / m^{2} \\ ε_{0} & = & 8.85 \times 10^{- 12} C^{2} / N . m^{2} \\ e & = & 1.6 \times 10^{- 19} C \\ r & = & 2.50 c m \\ W & = & \frac{(1.6 \times 10^{- 19} C) (2.50 c m) (10 μ C / m^{2})}{8.85 \times 10^{- 12} C^{2} / N . m^{2}} \\ = & - 4.5 \times 10^{- 15} J \end{array}$

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A 5.0-cm-diameter metal ball has a surface charge density of $10 m C / m 2$ . How much work is required to remove one electron from this ball?

Short Answer

Step by step solution

Definition of potential charge

theory of work energy

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A 5.0-cm-diameter metal ball has a surface charge density of 10mC/m2. How much work is required to remove one electron from this ball?

Short Answer

Step by step solution

Definition of potential charge

theory of work energy

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Kinematics Physics

Engineering Physics

Magnetism and Electromagnetic Induction

Famous Physicists

Quantum Physics

Nuclear Physics

Study anywhere. Anytime. Across all devices.

A 5.0-cm-diameter metal ball has a surface charge density of $10 m C / m 2$ . How much work is required to remove one electron from this ball?