3/21/20161 ELECTRICITY AND MAGNETISM Phy 220 Chapter2: Gauss’s Law

Electric Flux and Gauss’s Law A convenient technique was introduced by Karl F. Gauss ( ) to calculate electric fields. Requires symmetric charge distributions. Technique based on the notion of electrical flux. 3/21/20162

2.1 Electric Flux To introduce the notion of flux, consider, first, a situation where the electric field is uniform in magnitude and direction. Consider also that the field lines cross a surface of area A which is perpendicular to the field. The number of field lines per unit of area is constant. The flux, , is defined as the product of the field magnitude by the area crossed by the field lines. 3/21/20163 Area=A

2.1 Electric Flux Units: Nm 2 /C in SI units. Find the electric flux through the area A = 2 m 2, which is perpendicular to a uniform electric field of magnitude E=22 N/C 3/21/20164 Answer:  = 44 Nm 2 /C.

2.1 Electric Flux If the surface is not perpendicular to the field, the expression of the flux becomes: Where  is the angle between the field and a normal to the surface. 3/21/20165 Because the number of lines that go through the area A is the same as the number that go through A’, we conclude that the flux through A is equal to the flux through A’ which is given by

2.1 Electric Flux Remark: When an area is constructed such that a closed surface is formed, we shall adopt the convention that the flux of lines passing into the volume is negative and the flux of those passing out of the volume is positive. 3/21/20166 The flux through an area element can be positive (element 1), zero (element 2), or negative (element 3).

3/21/20167 If we define a vector whose magnitude equals the area of the surface and whose direction is perpendicular to the surface (directed out of it when it is a closed surface), equation [1] will give: 2.1 Electric Flux We assumed a uniform electric field in the preceding discussion. In more general situations, the electric field may vary over a surface. Therefore, the general definition of electric flux is: Equation [2] is a surface integral, which means it must be evaluated over the surface in question.

3/21/20168 EXAMPLE 1: Flux Through a Cube Consider a uniform electric field oriented in the x-direction. Find the electric flux through each surface of a cube with edges L oriented as shown in Figure, and the net flux.

3/21/ Electric Field of a Continuous Charge Distribution Charge Distribution The electric field at P due to one charge element carrying a charge Δq is How to calculate The electric field at P due to a continuous charge distribution? is a unit vector directed from the element toward P. The total electric field at P due to all elements in the charge distribution is approximately Because the charge distribution is modeled as continuous, the total field at P in the limit Δq i → 0 is where the integration is over the entire charge distribution.

3/21/  When the electric charge is uniformly distributed on a line, on a surface, or throughout a volume, it is convenient to use the concept of a charge density along with the following notations: If a charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is defined (in units of coulombs per cubic meter (C/m 3 ).) by If a charge Q is uniformly distributed on a surface of area A, the surface charge density σ is defined (in units of coulombs per square meter (C/m 2 ).) by If a charge Q is uniformly distributed along a line of length l, the linear charge density λ is defined (in units of coulombs per meter (C/m).) by

3/21/ If the charge is nonuniformly distributed over a volume, surface, or line, the amounts of charge dq in a small volume, surface, or length element are

2.3 Gauss’s Law The net flux passing through a closed surface surrounding a charge Q is proportional to the magnitude of Q: In free space, the constant of proportionality is 1/  o where  o is called the permittivity of of free space. 3/21/201612

2.3 Gauss’s Law The net flux passing through any closed surface is equal to the net charge inside the surface divided by  o. 3/21/201613

3/21/ Consider a positive point charge q surrounded by a spherical surface of radius r centered on the charge, as in Figure The magnitude of the electric field everywhere on the surface of the sphere is Note that the electric field is perpendicular to the spherical surface at all points on the surface. The electric flux through the surface is therefore EA, where A = 4πr 2 is the surface area of the sphere: This result says that the electric flux through a sphere that surrounds a charge q is equal to the charge divided by the constant ɛ 0. Using calculus, this result can be proven for any closed surface that surrounds the charge q.  This result can be generalized to any charge distribution.

3/21/ Example 2: Flux Due to a Point Charge

3/21/ Application of Gauss’s Law to Various Charge Distributions Gauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. In choosing the Gaussian surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the surface integral [4]. The Gaussian surface should satisfy one or more of the following conditions: 1.The magnitude of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product in Equation [4] can be expressed as a simple algebraic product E dA because the two vectors are parallel. 3. The dot product in Equation [4] is zero because the two vectors are perpendicular. 4. The field can be argued to be zero over the surface.

3/21/ A Spherically Symmetric Charge Distribution

3/21/ A Spherically Symmetric Charge Distribution

3/21/ A Spherically Symmetric Charge Distribution A plot of E versus r for a uniformly charged insulating sphere.

3/21/ A Cylindrically Symmetric Charge Distribution

3/21/ A Cylindrically Symmetric Charge Distribution  For points close to a finite line charge and far from the ends, the precedent result gives a good approximation of the value of the field.

3/21/ A Plane of charge Solution By symmetry, must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane. The direction of on one side of the plane must be opposite its direction on the other side, as shown in Figure. The gaussian surface is a small cylinder whose axis is perpendicular to the plane and whose ends each have an area A 0 and are equidistant from the plane. No electric field lines pass through the curved surface of the cylinder, only through the two ends, which have total area 2A 0

3/21/ A Plane of charge