Physics Chapter 1: Electric Charges and Fields

An electric dipole is a pair of equal and opposite charges separated by a small distance.

Electric flux is the measure of the electric field passing through a given surface. Its SI unit is Volt-meter (Vm).

Coulomb's law holds true for point charges at rest. It is not valid for charges in motion or for charges distributed continuously.

<p>Electric dipole moment is the product of the magnitude of one of the charges and the distance between the two charges. Its SI unit is Coulomb-meter (Cm).</p> Signup and view all the answers

<p>Gauss's theorem states that the total electric flux through a closed surface is proportional to the enclosed electric charge.</p> Signup and view all the answers

<p>Electric flux is the measure of the electric field passing through a given surface. There are two main types of flux: electric flux and magnetic flux.</p> Signup and view all the answers

<p>Coulomb's inverse square law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Unit charge is defined as the charge that exerts a force of 1 Newton on an identical charge placed 1 meter away from it. The law applies to point charges at rest and is valid for distances much larger than the size of the charges.</p> Signup and view all the answers

<p>Intensity of electric field at a point is defined as the force experienced by a unit positive charge placed at that point. Its unit is Newton per Coulomb (N/C).</p> Signup and view all the answers

<p>Properties of electric lines of force: 1) Start from positive charges and end at negative charges. 2) They never intersect each other. 3) The tangent at any point on a line of force gives the direction of the electric field at that point. 4) They are closer together where the electric field is stronger. Diagram of an electric lines of force of an isolated charge will include radial outward lines emanating from a positive charge, with each line being perpendicular to the surface of the charge.</p> Signup and view all the answers

<p>When a surface is parallel to an electric field, the angle between the electric field vector and the area vector of the surface is 90 degrees. Since the cosine of 90 degrees is zero, the electric flux through the surface is also zero. This is because the electric field lines do not pass through the surface.</p> Signup and view all the answers

<p>Consider an electric dipole with charges +q and -q separated by a distance 2l. Let P be a point on the axial line at a distance r from the center of the dipole. The electric field intensity due to the positive charge is E1 = kq/(r - l)^2, and the eletric field intensity due to the negative charge is E2 = kq/(r + l)^2. Thus, the net electric field intensity at P is E = E1 - E2. Simplifying this expression gives E = (2kpql)/(r^3) where p = 2ql is the electric dipole moment.</p> Signup and view all the answers

<p>Let P be a point on the equatorial line of the dipole at a distance r from the center of the dipole. The electric field intensity at P due to the positive charge Q is E1 = kQ/[(r^2) + (l^2)]^1/2, and the electric field intensity due to the negative charge -Q is E2 = kQ/[(r^2) + (l^2)]^1/2. The net electric field intensity at P is E = 2E1*sinθ where θ is the angle between the electric field vector and the line joining the charge to the point P. Simplifying this expression gives E = kpq/[(r^2) + (l^2)]^3/2 where p is the electric dipole moment.</p> Signup and view all the answers

<p>Gauss's theorem states that the total electric flux through a closed surface is proportional to the enclosed electric charge. Proof: Consider a closed surface containing a charge Q. Divide the surface into small areas ∆S, each with unit outward normal vector n̂. The flux through each of these areas is given by ∆Φ = E⋅∆S = E∆S cos θ where E is the electric field intensity at ∆S and θ is the angle between E and n̂. The total flux through the surface is then Φ = ∑ ∆Φ = ∑ E∆S cos θ. Using Coulomb's law, we can write E = kQ/r^2. Substituting into the equation and simplifying, we find that the total flux is Φ = 4πkQ. Since k = 1/(4πεο), the flux can be re-written as Φ = Q/εο. This proves that the total electric flux through a closed surface is proportional to the enclosed charge.</p> Signup and view all the answers

<p>Gauss's theorem: The total electric flux through a closed surface is proportional to the enclosed electric charge. Derivation of Coulomb's inverse square law using Gauss's theorem involves considering a spherical Gaussian surface centered on a point charge Q. The electric field is radial and uniform over the Gaussian surface, making the flux calculation straightforward. By applying Gauss's theorem and simplifying the flux equation, we arrive at Coulomb's inverse square law relating the force between charges to their magnitudes and the distance between them.</p> Signup and view all the answers