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What will the surface charge density be if the radius of the disk is doubledA circular disk has surface charge density 30 nCcm2.?
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How is the charge distributed on a non-spherical conductor?
Charge density would be more where the curvature is more. So pointed surface would have max charge density. Hence there is a chance of electrical discharge at the sharp points. This is known as Corona Discharge or Action of Points
What is the charge density for a conductor?
The charge density inside a conductor is always zero
Why is electric field inside a ring is zero?
If you refer to Gauss's law, it states the electric flux through any closed surface is proportionate to the enclosed electric charge. The electric flux density is the same as the electric field intensity. A Gaussian surface is a closed, three dimensional surface (there's no holes in it). Here's an example to help clarify what this is and is not saying. Suppose I have a clear glass ball. The "light charge" inside the ball is zero because there is no light source inside the ball. If I put the ball in the sunlight, light will go into one side, and out the other (ignore any sort of prism effect, etc. just don't think too hard about this example!). This ball still does not have an internal "light charge" because the light flowing into the ball is equivalent to the light flowing out (the "light density" through the surface sums to zero, or the line integral of the light density = 0 for this surface). If I put a light source inside the ball, the line integral of "light density" leaving the ball would be proportional to the "light charge" inside the ball; in other words the line integral tells you what is enclosed by the Gaussian surface (my fictitious light source, but not the sun). Even if I put it in the sunlight again, the line integral will remove the "light charge" due to the sun and I will be left with only my internal light source. In both these instances, absolutely nothing is being stated about the "light density" / "light intensity" inside the ball. For both instances, there is a light intensity INSIDE the ball, even though the "light charge" inside is non zero in only one case. Relating to the question, this means if you have a Gaussian surface (such as a sphere), and it has/does not have an enclosed electric charge, you can have an electric field through the sphere - the fact this field is there tells you nothing about the internal charge of the Gaussian surface until you perform the line integral to measure what's coming in and what's going out. So, what I'm stating is the question is not true - the electric field is not necessarily zero inside a Gaussian surface, even if the surface does not contain an electrically charged particle. This should be easily seen by taking the typical point charge example: You have a point charge, and you draw the Gaussian surface around it. The point charge radiates electric field lines in all directions away from itself. If you move the Gaussian surface to the left until the point charge is no longer enclosed in it, you will see the radiating electric field lines due to this point charge still go into and out of the surface (so there is an electric field due to the point charge inside the surface), but the point charge is no longer enclosed by the surface (so the line integral sums to zero).
What is the conservation of charge law from maxwell's equations?
The conservation of charge law from Maxwell's equations states that the current through any enclosed surface is equal to the time rate of charge within the surface.
What happens to the charge on the conductive sphere when it is connected to a source of charge?
The charge on the conductive sphere spreads out uniformly over the surface of the sphere.
The relative distribution of charge density on the surface of a conducting solid depends on what?
the density of the conductor
What is the Difference between line charge density and surface charge density?
The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder.
An isolated conducting sphere whose radius R equals 1M has a charge 1.1nc the energy density at the surface of sphere?
The energy density at the surface of a charged conductor is the surface charge density squared , divided by 2 x the permittivity of free space. The surface charge density is the charge divided by the area it sits on. So if, e = permittivity = 8.85 x 10^-12 CC/Nmm and D = surface charge density, and U = energy density and R = radius of sphere and q = charge on sphere, then; U = (1/2e) x D^2 where D = q/4piR^2 = 1.1 x 10^-9/(4 x 3.14 x 1) = 8.76 x 10^-11 , where 4piR^2 is the surface area of a sphere. So; D^2 = 76.7 x 10^-22 then ; U = (76.7 x 10^-22)/(17.7 x 10^-12) = 4.33 x 10^-10 Joules/mmm
What is Charge Density?
In electromagnetism, charge density is a measure of electric charge per unit volume of space, in one, two or three dimensions. More specifically: the linear, surface, or volume charge density is the amount of electric charge per unitlength, surface area, or volume, respectively. The respective SI units are C·m−1, C·m−2 or C·m−3.[1]Like any density, charge density can depend on position, but because charge can be negative - so can the density. It should not be confused with the charge carrier density, the number of charge carriers (e.g. electrons, ions) in a material per unit volume, not including the actual charge on the carriers.In chemistry, it can refer to the charge distribution over the volume of a particle; such as a molecule, atom or ion. Therefore, a lithium cation will carry a higher charge density than a sodium cation due to the lithium cation's having a smaller ionic radius, even though sodium has more electrons (11) than lithium (3).
How is the charge distributed on a non-spherical conductor?
Charge density would be more where the curvature is more. So pointed surface would have max charge density. Hence there is a chance of electrical discharge at the sharp points. This is known as Corona Discharge or Action of Points
What has the author David Kenneth Davies written?
David Kenneth Davies has written: 'The absolute determination of surface charge density'
What is the charge density for a conductor?
The charge density inside a conductor is always zero
What quality must the charge density on the surface of a conducting wire possess if an electric field is to act on the negatively charged electrons inside the wire?
non uniform
Why is electric field inside a ring is zero?
If you refer to Gauss's law, it states the electric flux through any closed surface is proportionate to the enclosed electric charge. The electric flux density is the same as the electric field intensity. A Gaussian surface is a closed, three dimensional surface (there's no holes in it). Here's an example to help clarify what this is and is not saying. Suppose I have a clear glass ball. The "light charge" inside the ball is zero because there is no light source inside the ball. If I put the ball in the sunlight, light will go into one side, and out the other (ignore any sort of prism effect, etc. just don't think too hard about this example!). This ball still does not have an internal "light charge" because the light flowing into the ball is equivalent to the light flowing out (the "light density" through the surface sums to zero, or the line integral of the light density = 0 for this surface). If I put a light source inside the ball, the line integral of "light density" leaving the ball would be proportional to the "light charge" inside the ball; in other words the line integral tells you what is enclosed by the Gaussian surface (my fictitious light source, but not the sun). Even if I put it in the sunlight again, the line integral will remove the "light charge" due to the sun and I will be left with only my internal light source. In both these instances, absolutely nothing is being stated about the "light density" / "light intensity" inside the ball. For both instances, there is a light intensity INSIDE the ball, even though the "light charge" inside is non zero in only one case. Relating to the question, this means if you have a Gaussian surface (such as a sphere), and it has/does not have an enclosed electric charge, you can have an electric field through the sphere - the fact this field is there tells you nothing about the internal charge of the Gaussian surface until you perform the line integral to measure what's coming in and what's going out. So, what I'm stating is the question is not true - the electric field is not necessarily zero inside a Gaussian surface, even if the surface does not contain an electrically charged particle. This should be easily seen by taking the typical point charge example: You have a point charge, and you draw the Gaussian surface around it. The point charge radiates electric field lines in all directions away from itself. If you move the Gaussian surface to the left until the point charge is no longer enclosed in it, you will see the radiating electric field lines due to this point charge still go into and out of the surface (so there is an electric field due to the point charge inside the surface), but the point charge is no longer enclosed by the surface (so the line integral sums to zero).
What must be used in order to separate chemicals in a compound?
Richard Bader is a scientist from McMaster University who developed a way of dividing molecules into atoms. He uses zero flux surfaces to divide atoms. (A zero flux surface is a 2-D surface on which the charge density is a minimum perpendicular to the surface). Usually in molecular systems, the charge density reaches a minimum between atoms and this is a natural place to separate atoms from each other. more info- http://theory.cm.utexas.edu/bader/
How can you calculate the total charge on a spherical body if its radius R and surface charge density sigma are given?
Uh, assuming by "spherical body" you mean "spherical shell of insignificant thickness" then it would just be 4*pi*(R^2)*sigma
What is the unit of volumetric density?
Volumetric density is the density based upon the volume of an object.
