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Maharashtra State BoardHSC Science (Computer Science) 12th Standard Board Exam
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अभ्यासक्रम

Applications of Gauss' Theorem

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Topics

Estimated time: 4 minutes
  • Electric Field due to a Point Charge
  • Electric field due to an Infinite Line of Charge 
  • Electric Field due to an Infinite Plane Sheet of Charge
  • Electric Field due to Two Infinite Parallel Sheets of Charge
  • Electric Field Intensity Just Outside a Charged Conductor
  • Electric Field due to a Uniformly Charged Thin Spherical Shell
  • Electric Field due to a Uniformly Charged Sphere (Optional)
CBSE: Class 12

Key Points: Applications of Gauss' Theorem

  • Gauss’s law is useful for finding the electric field in highly symmetric charge distributions (line, plane, sphere).
  • For an infinitely long charged wire, the electric field is radial and depends only on the distance r from the wire.
  • Electric field due to an infinite line charge decreases with distance:
    E = \[\frac{\lambda}{2\pi\varepsilon_0r}\]
  • For an infinite plane sheet, the electric field is uniform and does not change with distance.
  • Electric field due to an infinite plane sheet is:
    E = \[\frac{\sigma}{2\varepsilon_0}\]
  • For a uniformly charged spherical shell, the field outside behaves like a point charge at the centre:
    E = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\]
  • Inside a uniformly charged spherical shell, the electric field is zero.
 

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Related QuestionsVIEW ALL [20]

  1. Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it.
  2. Draw a graph showing the variation of electric field intensity E with r, for r > R and r < R.

"For any charge configuration, equipotential surface through a point is normal to the electric field." Justify.

A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown the figure. Use Gauss' law to obtain the expression for the electric flux through the square.

A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.

  1. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
  2. What is the value of the point charge?

An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre, and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of the magnitudes of its charge and linear charge density λ on the wire.

Consider a sphere of radius R with charge density distributed as

ρ(r)  = kr for r ≤ R

        = 0 for r > R

  1. Find the electric field at all points r.
  2. Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

State Gauss’ Law.

Explain why, for a charge configuration, the equipotential surface through a point is normal to the electric field at that point

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