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Chapter 19: Problem 3

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e )\) moves from the outside to the inside?

Short Answer

Expert verified
The work done is \(-1.12 \times 10^{-20}\) Joules.

Step by step solution

01

Understand the Electric Potential Difference

The electric potential difference between two points is the work done per unit charge to move a charge between these points. Here, the potential is higher outside the cell by 0.070 V compared to inside.
02

Identify the Charge of the Sodium Ion

A sodium ion has a charge equal to the elementary charge \( e \), which is approximately \( 1.6 \times 10^{-19} \) coulombs.
03

Calculate the Work Done by Electric Force

The work done \( W \) by the electric force is given by the formula: \( W = q \Delta V \), where \( q \) is the charge and \( \Delta V \) is the electric potential difference. Here, \( q = +e \) and \( \Delta V = -0.070 \text{ V} \) (since potential decreases going inside the cell).
04

Substitute in the Values

Substitute \( q = 1.6 \times 10^{-19} \) C and \( \Delta V = -0.070 \text{ V} \) into the formula: \[ W = (1.6 \times 10^{-19}) \times (-0.070) \]
05

Calculate the Result

Perform the multiplication: \[ W = -1.12 \times 10^{-20} \text{ Joules} \]. The negative sign indicates the work done by the electric force is in the direction of the force, from outside to inside the cell.

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Key Concepts

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

Work Done by Electric Force
The work done by an electric force is a key concept in understanding how charges interact in electric fields. When a charged particle, such as a sodium ion, moves in an electric field, work is done by the electrical forces acting on it. This work can be calculated by multiplying the charge of the particle by the electric potential difference it moves through.

In the context of our exercise, the electric potential difference (\( \Delta V \) \between the outside and inside of the cell is given as 0.070 V. The essential formula for calculating work done by the electric force is:
  • \( W = q \times \Delta V \)
where \( q \) is the charge and \( \Delta V \) \is the electric potential difference. When plugging the values into the formula, the work done comes out as negative. This simply indicates that the work is done by the electric field in favor of the sodium ion's movement, from a higher to a lower potential.
Sodium Ion Charge
Sodium ions play a crucial role in cellular activities, particularly in nerve signaling and muscle contractions. Each sodium ion bears a single positive elementary charge, making it \( \positive. \)

A sodium ion is represented by the chemical formula \( \\text{Na}^+ \), indicating that it has lost one electron compared to its neutral state. This results in a positive charge equal to one elementary charge.

In numerical terms, a sodium ion's charge is \( \+e \), which corresponds to approximately 1.6 \( \\times \10^{-19} \) coulombs. This is a small amount of charge, but it is significant in biological systems due to the large number of ions involved. Understanding the charge of ions like sodium helps in calculating how these ions move across cellular membranes, influenced by electrical fields.
Elementary Charge
The elementary charge is a fundamental and indivisible unit of electric charge. It is essential to our understanding of how charged particles behave in electric fields and is the basis for more complex calculations.

The elementary charge, denoted by \( \e\), is approximately \( 1.6 \times 10^{-19} \) coulombs. It is the charge carried by a single proton (positive) or, oppositely, the charge of a single electron (negative).

This minute yet crucial value allows scientists to quantify the electrical interactions of atoms and subatomic particles. When dealing with ions, like \( \text{Na}^+ \), it provides a base measure of the charge they possess. Calculations involving the elementary charge are foundational in physics and chemistry for understanding electric forces and their effects on materials and living systems.

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Most popular questions from this chapter

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that the electric field can have without breakdown occurring. The dielectric strength of air is \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m},\) and that of neoprene rubber is \(1.2 \times 10^{7} \mathrm{V} / \mathrm{m} .\) A certain air-gap, parallel plate capacitor can store no more than 0.075 \(\mathrm{J}\) of electrical energy before breaking down. How much energy can this capacitor store with- out breaking down after the gap between its plates is filled with neoprene rubber?

A particle has a charge of \(+1.5 \mu \mathrm{C}\) and moves from point \(A\) to point \(B,\) a distance of 0.20 \(\mathrm{m}\) . The particle experiences a constant elec- tric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at \(A\) and at \(B\) is EPE \(_{A}-\mathrm{EPE}_{B}=+9.0 \times 10^{-4} \mathrm{J}\) . (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

Two capacitors are identical, except that one is empty and the other is filled with a dielectric \((\kappa=4.50)\) . The empty capacitor is conected to a \(12.0-\mathrm{V}\) battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\) , arriving there with a speed of 42 \(\mathrm{m} / \mathrm{s}\) . The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store 11\(\mu \mathrm{C}\) of charge and \(5.0 \times 10^{-5} \mathrm{J}\) of energy. When used with capacitor \(\mathrm{B}\) , which has a capacitance of 6.7\(\mu \mathrm{F}\) , this voltage causes the capacitor to store a charge that has a magnitude of \(q_{\mathrm{B}}\) Determine \(q_{\mathrm{B}}\) .
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