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A point charge \(q\) produces an electric field of magnitude \(2 \mathrm{NC}^{-1}\) at a point distance \(0.25 \mathrm{~m}\) from it. What is the value of charge? (a) \(139 \times 10^{-11} \mathrm{C}\) (b) \(139 \times 10^{11} \mathrm{C}\) (c) \(13.9 \times 10^{-11} \mathrm{C}\) (d) \(1.39 \times 10^{11} \mathrm{C}\)

Short Answer

Expert verified
The charge is \(139 \times 10^{-11} \, \mathrm{C}\), option (a).

Step by step solution

01

Understand the formula for electric field due to a point charge

The formula for the electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by \[ E = \frac{k \, |q|}{r^2} \]where \(k\) is Coulomb's constant with a value of approximately \(8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2\).
02

Plug in the known values

Given that the electric field \(E\) is \(2 \, \mathrm{NC}^{-1}\) and the distance \(r\) is \(0.25 \, \mathrm{m}\), substitute these values into the formula:\[ 2 \, \mathrm{NC}^{-1} = \frac{8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2 \cdot |q|}{(0.25 \, \mathrm{m})^2} \]
03

Solve for the charge \(q\)

Rearrange the equation to solve for \(|q|\):\[ |q| = \frac{2 \, \mathrm{NC}^{-1} \, \times (0.25 \, \mathrm{m})^2}{8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2} \]\[ |q| = \frac{2 \, \times 0.0625}{8.99 \times 10^9} \]\[ |q| = \frac{0.125}{8.99 \times 10^9} \]
04

Calculate the value

Now, perform the calculation:\[ |q| = \frac{0.125}{8.99 \times 10^9} \approx 1.39 \times 10^{-11} \, \mathrm{C} \]
05

Select the correct option

Comparing the calculated charge \(1.39 \times 10^{-11} \, \mathrm{C}\) with the given options, the correct answer corresponds to option (a): \(139 \times 10^{-11} \, \mathrm{C}\).

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Key Concepts

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

Coulomb's constant
Coulomb's constant, denoted as \(k\), is a fundamental constant in physics that quantifies the amount of force between two charges separated by a distance in a vacuum. It's derived from the principle known as Coulomb's Law, which defines the force between point charges.

The value of Coulomb's constant is approximately \(8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2\). This value is crucial because it helps us calculate the electric force and electric field between charges.
  • It provides the scaling factor for the strength of the electric field generated by a point charge.
  • It ensures consistency in calculations involving electric forces and fields across different problems.
Understanding this constant is vital for anyone working with electric fields because it plays a key role in determining the forces and fields due to charged particles.
Electric field calculation
The electric field calculation involves determining the effect of a charged particle on its surroundings, specifically when we have a point charge. The electric field \(E\) at a point in space is influenced by a source charge \(q\) and the distance \(r\) from that charge.

The formula to calculate the electric field due to a point charge is given by:
\[ E = \frac{k \cdot |q|}{r^2} \]
where:
  • \(E\) is the electric field magnitude.
  • \(k\) is Coulomb's constant.
  • \(|q|\) is the absolute value of the charge you're considering.
  • \(r\) is the distance from the charge to the point where the field is being calculated.
This equation shows how the electric field strength decreases with the square of the distance from the charge. Hence, it's an inverse square law, meaning doubling the distance results in a field strength that's only one-fourth as strong. Calculating the electric field accurately helps in understanding how charged objects influence their environment.
Point charge
A point charge is an idealized model of a charged particle where the charge is assumed to be concentrated at a single point in space. This simplifies the mathematics when calculating electric fields and forces.

Point charges are commonly used in electrostatics to model real-world charged objects, especially when those objects are small compared to the distances involved in the problem. When working with point charges, it's important to consider the effect on nearby objects and fields.
  • They generate electric fields that radiate outward from the charge.
  • The field strength diminishes with distance, following the inverse square law.
  • Calculations involving point charges help understand fundamental interactions in electrostatics.
Using point charges in calculations provides valuable insights into the behavior of charged systems and aids in designing devices like capacitors, sensors, and other electronics.

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Most popular questions from this chapter

The maximum field intensity on the axis of a uniformly charged ring of charge \(q\) and radius \(R\) will be (a) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{3 \sqrt{3} R^{2}}\) (b) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 q}{3 R^{2}}\) (c) \(\frac{1}{4 \pi E_{0}} \cdot \frac{2 q}{3 \sqrt{3} R^{2}}\) (d) \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 q}{2 \sqrt{3} R^{2}}\)

A charged particle of mass \(m\) and charge \(q\) is released from rest in an electric field of constant magnitude \(E\). The kinetic energy of the particle after time \(t\) is (a) \(\frac{E^{2} q^{2} t^{2}}{2 m}\) (b) \(\frac{2 E^{2} t^{2}}{q m}\) (c) \(\frac{\mathrm{Eqm}}{2 t}\) (d) \(\frac{E q^{2} m}{2 t^{2}}\)

A regular hexagon of side \(10 \mathrm{~cm}\) has a charge \(5 \mu \mathrm{C}\) at each of its vartices. The potential at the centre of the hexagon is? [NCERT] (a) \(3.7 \times 10^{6} \mathrm{~V}\) (b) \(2.7 \times 10^{6} \mathrm{~V}\) (c) \(4 \times 10^{6} \mathrm{~V}\) (d) \(5 \times 10^{7} \mathrm{~V}\)

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be (a) \(100 \mathrm{~N}\) (b) \(121 \mathrm{~N}\) (c) \(99 \mathrm{~N}\) (d) None of these

A \(600 \mathrm{pF}\) capacitor is charged by a \(200 \mathrm{~V}\) supply. Then, it is disconnected from the supply and is connected to another uncharged \(600 \mathrm{pF}\) capacitor. How much electrostatic energy is lost in the process? [NCERT] (a) \(4 \times 10^{-6} \mathrm{~J}\) (b) \(6 \times 10^{-6} \mathrm{~J}\) (c) \(5 \times 10^{-6} \mathrm{~J}\) (d) \(8 \times 10^{-6} \mathrm{~J}\)

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