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Chapter 20: Problem 25

A 68 -nC charge experiences a 150 -mN force in a certain electric field. Find (a) the field strength and (b) the force that a \(35-\mu \mathrm{C}\) charge would experience in the same field.

Short Answer

Expert verified
a) The field strength is \(22.06 * 10^5 \, N/C\); b) The force that a \(35-\mu C\) charge would experience is 7.72 N.

Step by step solution

01

Calculate Electric Field Strength

The electric field strength (E) can be calculated using the given force (F) and the charge (q), using the formula \(E = F/q\). Make sure to convert the force from millinewtons (mN) to newtons (N) and the charge from nanocoulombs (nC) to coulombs (C) by multiplying by \(10^-3\) and \(10^-9\) respectively, this yields \(E = 150*10^-3/\(68*10^-9 = 22.06 * 10^5 \, N/C\)
02

Calculate Force

Now use the calculated electric field strength to find the force that a \(35-\mu C\) charge would experience in the same field. Use the formula \(F = qE\). Remember to convert the charge from microcoulombs (\(\mu C\)) to coulombs (C) by multiplying by \(10^-6\), then multiply this value by the electric field to calculate the force. Thus \(F = 35*10^-6 * 22.06 * 10^5 \, N/C = 7.72 \, N\)

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Key Concepts

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

Charge Conversion
Understanding charge conversion is essential in electric calculations. Charges are often given in units like nanocoulombs (nC) or microcoulombs (渭C), which need to be converted to coulombs (C) for calculations.
1 Coulomb is the base SI unit, and converters are used to shift between different scales:
  • To convert from nanocoulombs to coulombs, multiply by 10鈦烩伖.
  • To convert from microcoulombs to coulombs, multiply by 10鈦烩伓.
In our exercise, we started with a 68-nC charge, which is converted to 0.068 渭C or 68脳10鈦烩伖 C for the electric field strength calculation.
Larger charges, like the 35-渭C, are converted similarly: 35脳10鈦烩伓 C. These simple conversion rules ensure the correct unit usage in further calculations, keeping computations accurate.
Electric Force Calculation
Electric force denotes the influence of an electric field on a charge. It can be calculated using the formula:
\[ F = qE \]
where:
  • \( F \) represents the electric force in newtons (N).
  • \( q \) is the charge in coulombs (C).
  • \( E \) is the electric field strength in newtons per coulomb (N/C).
The exercise involved calculating the force exerted on a 35-渭C charge in an identified electric field strength of 22.06脳10鈦� N/C.
Following conversion, the charge input is 35脳10鈦烩伓 C, which is then multiplied by the field strength, resulting in a force of 7.72 N.
This method highlights how electric forces change with different charges and field strengths.
Coulomb's Law
Coulomb's Law is a fundamental principle describing the electric force between two point charges. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In its formula form, Coulomb's Law is expressed as:
\[ F = k \times \frac{{|q_1 \times q_2|}}{{r^2}} \]
where:
  • \( F \) is the electric force.
  • \( k \) is Coulomb's constant (approximately 8.99脳10鈦� N路m虏/C虏).
  • \( q_1 \) and \( q_2 \) represent the magnitudes of the charges.
  • \( r \) is the distance separating the charges.
In this given exercise, two steps derive individual outputs making use of electric field strength principles rather than direct interaction between two charges.
However, understanding Coulomb's Law is crucial for context, as it paves the way for assessing how electric fields originate from point charges.

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Most popular questions from this chapter

A thin rod extends along the \(x\) -axis from \(x=0\) to \(x=L\) and carries line charge density \(\lambda=\lambda_{0}(x / L)^{2},\) where \(\lambda_{0}\) is a constant. Find the electric field at \(x=-L\)

Earth carries a net charge of about \(-5 \times 10^{5} \mathrm{C}\). How many more electrons are there than protons on Earth?

A straight wire 10 m long carries \(25 \mu C\) distributed uniformly over its length. (a) What's the line charge density on the wire? Find the electric field strength (b) \(15 \mathrm{cm}\) from the wire axis, not near either end, and (c) 350 m from the wire. Make suitable approximations in both cases.

Why does a dipole, which has no net charge, produce an electric field?

A \(65-\mu \mathrm{C}\) point charge is at the origin. Find the electric field at the points (a) \(x=50 \mathrm{cm}, y=0 \mathrm{cm} ;\) (b) \(x=50 \mathrm{cm}, y=50 \mathrm{cm}\) and \((\mathrm{c}) x=25 \mathrm{cm}, y=-75 \mathrm{cm}\)
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