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Questions and Answers

What is an electric field?

The region around a charge in which it can exert the force of attraction or repulsion on other charged bodies.

What does electric field intensity represent?

The electrostatic force on a positive charge at a specific field point.

What is the S.I. unit of electric field?

N/C or Vm-1

How is electric field intensity represented mathematically?

<p>$E = \frac{F}{q_0}$</p> Signup and view all the answers

What is the magnitude of the electric field that accelerates an electron at 1.84 x 10^{10} m/s^2?

<p>0.01 N/C</p> Signup and view all the answers

What is the electric force on an electron in an electric field of 3.0 x 10^6 N/C?

<p>4.8 x 10^{-13} N</p> Signup and view all the answers

What electric field magnitude will balance the weight of an alpha particle with mass 6.64 x 10^{-27} kg and charge 2e?

<p>2.03 x 10^{5} N/C</p> Signup and view all the answers

What is the ratio of electric force to gravitational force for a proton under an electric field of 1.5 x 10^3 N/C?

<p>1.5 x 10^{11}</p> Signup and view all the answers

What would be the acceleration experienced by a proton in an electric field of 2.16 x 10^4 N/C?

<p>2.07 x 10^{12} m/s^2</p> Signup and view all the answers

If a particle with charge -2.0 x 10^{-9} C experiences a downward force of 3.0 x 10^{-6} N, what is the magnitude of the electric field?

<p>1.5 x 10^3 N/C</p> Signup and view all the answers

Study Notes

Electric Field

  • The region around a charged object where it can exert an electric force on other charged bodies is called an electric field.

Electric Field Intensity

  • The electric field intensity at a point is the electrostatic force exerted on a positive test charge at that point.
  • It is defined as the force per unit test charge:
    • E = F / q0
  • The test charge q0 must be infinitesimally small to avoid disturbing the field.
  • The unit of electric field intensity is Newtons per Coulomb (N/C) or Volts per meter (V/m).

Worked Examples

  • Problem 28-1: An electron is accelerated eastward by an electric field with an acceleration of 1.84 x 10^10 m/s². The magnitude of the electric field is calculated by equating the electric force (F = qE) with the force due to acceleration (F = ma). Solving for E gives 0.01 N/C.
  • Problem 28-2: Humid air ionizes in an electric field of 3.0 x 10^6 N/C. The electric force on an electron and on an ion (missing a single electron), both in the field, is calculated using F = qE. In both cases, the force is 4.8 x 10^-13 N.
  • Problem 28-3: An alpha particle with mass 6.64 x 10^-27 kg and charge 2e (twice the charge of an electron) is balanced against gravity by an electric field. The magnitude of the electric field is calculated by equating the electric force (F = qE) and the gravitational force (F = mg), resulting in E = 2.03 x 10^5 N/C. The field must be directed upward to oppose the downward gravitational force.
  • Problem 28-4: A particle with charge -2.0 x 10^-9 C experiences a downward force of 3.0 x 10^-6 N in a uniform electric field.
    • The magnitude of the electric field is determined by E = F / q, which gives 1.5 x 10^3 N/C.
    • The force on a proton placed in this field is calculated as F = qE, resulting in 2.4 x 10^-16 N upward since the proton has a positive charge.
    • The gravitational force on the proton is calculated as F = mg, giving 1.64 x 10^-26 N downward.
    • The ratio of the electric force to the gravitational force for the proton is approximately 1.5 x 10^11, highlighting the dominance of electric forces over gravitational forces at the atomic level.
  • Problem 28-3b: A proton accelerates in an electric field of 2.16 x 10^4 N/C.
    • The acceleration is calculated by equating F = qE and F = ma, resulting in 2.07 x 10^12 m/s².
    • The final speed of the proton after traveling 1.22 cm is determined by using the third equation of motion (v² = vi² + 2as) with initial velocity vi = 0. The final speed is calculated to be... (the calculation was not completed in the provided text).

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