The general form of Gauss's law describes how a charge...

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where $\epsilon=\kappa \epsilon_{0}$ is the permittivity of the material. (a) $\mathrm{A}$ sheet with charge $Q$ uniformly distributed over its area $A$ is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude $E=Q / 2 A \epsilon$. (b) Two large sheets of area $A$, carrying opposite charges of equal magnitude $Q,$ are a small distance $d$ apart. Show that they create uniform electric field in the space between them, with magnitude $E=Q / A \epsilon$ (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is $A \epsilon / d=\kappa A \epsilon_{0} / d$.

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is
$$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$
where $\epsilon=\kappa \epsilon_{0}$ is the permittivity of the material. (a) $\mathrm{A}$ sheet with charge $Q$ uniformly distributed over its area $A$ is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude $E=Q / 2 A \epsilon$. (b) Two large sheets of area $A$, carrying opposite charges of equal magnitude $Q,$ are a small distance
$d$ apart. Show that they create uniform electric field in the space between them, with magnitude $E=Q / A \epsilon$
(c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae.
(d) Show that the capacitance of the pair of plates is $A \epsilon / d=\kappa A \epsilon_{0} / d$.

Key Concepts

Gauss's Law

Gauss's Law relates the net electric flux through a closed surface to the charge enclosed by that surface. This principle is used to derive expressions for the electric field due to symmetric charge distributions and is formulated as the integral of the electric field over an enclosing surface equaling the enclosed charge divided by the permittivity.

Permittivity

Permittivity quantifies how a material responds to an electric field and affects the formation of electric fields within the medium. It is typically expressed as ? = ???, where ?? is the permittivity of free space and ? is the dielectric constant, indicating how much the material can 'permit' the electric field compared to a vacuum.

Electric Field from Charge Distributions

This concept involves calculating the electric field produced by a particular distribution of charge. Using Gauss's Law, one can determine the field generated by configurations such as an infinite plane of charge, where symmetry simplifies the computation.

Uniform Electric Field

A uniform electric field has constant magnitude and direction within a certain region. This condition is often realized between parallel plates (or sheets) of charge, where the field between the plates is uniform due to the symmetry and distribution of the charges.

Potential Difference and its Relation to the Electric Field

The potential difference between two points in space is related to the electric field by the line integral of the field along the path connecting the points. This relationship is essential in linking the work done on or by charges moving in an electric field to differences in electric potential.

Capacitance

Capacitance is defined as the ability of a system to store charge per unit potential difference. For parallel plate configurations, capacitance is determined by the plate area, the separation distance, and the permittivity of the intervening dielectric, encapsulating how geometry and material properties influence charge storage.

Transcript

00:01 Hello everyone this is a problem based on gosla which is described for vacuum surface integral of electric field for the closed hypothetical surface to be total charge inside it divided by permittivity of free space and for material this is for vacuum or free space and for material it would be e.

00:42 Upon k a epsilon not here k is dielectric constant of the material in the first part we have to find the electric field due to charge q q distributed over area a surrounded by dielectric in second part we have to calculate the electric field between two parallel plates large sheet carrying equal and opposite charge each of area a.

02:49 See part.

03:04 Find the potential of positive plate if negative plate having the potential to be zero.

03:25 Capistence of the pair of plates, let us start solving it.

03:50 Starting with part a, this is seat having the charge q and area a.

04:12 If we consider a pillar box of area a, which is much less than area of the seat, then charge inside it will be this portion, sigma into a dash and sigma having the value charge per in vit area that is q upon a.

04:59 Now if we apply the gosuram for electric field, suppose we are finding the field in this direction at this point.

05:10 E .d .a.

05:14 2b, q upon epsilon.

05:18 Here, epsilon is permittivity of medium, which is defined as epsilon not into k.

05:38 Now take the ground, it having the normal vector in this direction, this surface having the normal vector in this direction...

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