# A solid conducting sphere of radius a

Non-conducting supports are used to maintain this configuration of spheres. Take electric potential to be zero at infinity.

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## A solid conducting sphere of radius a

None of the above. A spherical conducting shell of inner radius r 1 and outer radius r 2 has a charge Q. What is the surface charge density on the inner and outer surfaces of the shell? A conducting spherical shell having inner radius a and outer radius b carries a net charge Q. If a point charge q is placed at the centre of this shell, then the surface charge density on the outer surface of the shell is given as:. A metallic spherical shell has an inner radius R 1 and outer radius R 2. A charge is placed at the centre of the spherical cavity. The surface charge density on the inner surface is. A hollow conducting sphere of inner radius r and outer radius 2R is given charge Q as shown in figure, then the. A solid spherical conducting shell has inner radius a and outer radius 2a. What must the excess charge of the shell be in order for the charge density on the inner and outer surface of the shell to be exactly equal? A conducting sphere of radius R and a concentric thick spherical shell of inner radius 2R and outer radius 3R is shown in figure. Then charge on inner sphere is.

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Submitted by Dawn J. We will assign your question to a Numerade educator to answer. A solid conducting sphere of radius 2. A conducting spherical shell of inner radius 4. Find the electric field at the following radii from the center of this charge configuration.

It turns out that in situations that have certain symmetries spherical, cylindrical, or planar in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. In these systems, we can find a Gaussian surface S over which the electric field has constant magnitude. Note that these symmetries lead to the transformation of the flux integral into a product of the magnitude of the electric field and an appropriate area. The direction of the electric field at point P is obtained from the symmetry of the charge distribution and the type of charge in the distribution. Here is a summary of the steps we will follow:. They are. To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussian surface for that symmetry, applying the remaining four steps. A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in space and not on the direction. Thus, it is not the shape of the object but rather the shape of the charge distribution that determines whether or not a system has spherical symmetry. Figure 6.

### A solid conducting sphere of radius a

The electric field of a point charge Q can be obtained by a straightforward application of Gauss' law. Considering a Gaussian surface in the form of a sphere at radius r , the electric field has the same magnitude at every point of the sphere and is directed outward. The electric flux is then just the electric field times the area of the sphere. The electric field of a conducting sphere with charge Q can be obtained by a straightforward application of Gauss' law. The electric flux is then just the electric field times the area of the spherical surface. The electric field of a sphere of uniform charge density and total charge charge Q can be obtained by applying Gauss' law. The electric field inside a sphere of uniform charge is radially outward by symmetry , but a spherical Gaussian surface would enclose less than the total charge Q. The charge inside a radius r is given by the ratio of the volumes:. The spherically symmetric charge outside the radius r does not affect the electric field at r. It follows that inside a spherical shell of charge, you would have zero electric field.

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A hollow conducting sphere of inner radius r and outer radius 2R is given charge Q as shown in figure, then the. Was this answer helpful? Find a the charge distribution for the insulating sphere and the conducting spherical shell, and the magnitude of the electric field at the following distances from the center of the two spheres and shell: b 0. Rollercoaster Problems. Electric Charge. Phasors for Capacitors. Instant help, 24x7. Does the electric flux through the Heat Transfer. Ray Diagrams For Mirrors.

Imagine for a moment that we have two neutrally-charged but otherwise arbitrary conductors, separated in space. Figure 2. Clearly there is an electric field pointing out of the former, and into the latter, with the field lines leaving and landing perpendicular to the surfaces.

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