Well I doubt it would be able to hold a charge of that density in air. When a sphere of radius 1cm is charged to such a high level, the electric field will be far larger than the dielectric strength of air. Hence, a breakdown voltage will be achieved and electrons will flow even through air. It's the same principle that lightning works on. 1 coulomb is a huge amount of charge, and the surface charge density in this case will be too high for air to take. So, a sphere of radius 1 cm cannot hold 1 Coulomb of charge.
The charge all resides on the surface of the sphere, whether or not there's anything inside the surface. In principle, there's no limit on the amount of charge that can be jammed onto the sphere. The only limit is a practical one, that is, how much charge you can move and transfer to the sphere before it starts arcing back to the machinery or the support that's holding it.
When a sphere is cut into cross sections, the shape formed is a circle. This is because all cross sections of a sphere will be circular in shape, regardless of the angle or position of the cut.
OK, so I can't include MathML or images or TeX so I'm struggling to answer this question. Here's a verbal description of the equation V = ((PI^(n/2)) * r^n)/G((n/2)+1) where PI is 3.14159... G() is the Gamma Function (complex factorial) and ^ indicates "to the power of". Hope that helps. Source: Eqn 20 , p453 of Shannon's paper "Communication in the Presence of Noise"
A silo consists of a cylinder and a hemisphere. So, the volume of the silo is the volume of a cylinder plus the volume of the hemisphere.Volume of a sphere: 4/3πr3 ---> Hemisphere: 2/3πr3Volume of a cylinder: πr2hSo, combining the two equations, the full equation to finding the volume of a silo:V = 2/3πr3 + πr2h
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A metal sphere of radius 1 centimeter will not hold a charge of 1 coulomb. The electric field generated from the metal sphere of radiusÊat 1 centimeter will break down and neutralize any charges.
The surface charge density formula of a sphere is Q / 4r, where is the surface charge density, Q is the total charge on the sphere, and r is the radius of the sphere.
touch it with a charged object....
inner core
The formula for calculating the surface charge density of a sphere is: Q / 4r, where represents the surface charge density, Q is the total charge on the sphere, and r is the radius of the sphere.
The inner core of the Earth is a solid metallic sphere made primarily of iron and nickel with a radius of about 1216 kilometers. It is surrounded by the liquid outer core and the solid mantle.
A sphere of radius R surround a point charge Q, located at its center afind the electric flux ?
The radius of a sphere joins the center of the sphere to its circumference.
The radius of a sphere is equal distance from the center of the sphere to all points within the sphere.
bidyogammes
The formula for calculating the electric field of a sphere is E k Q / r2, where E is the electric field, k is the Coulomb's constant (8.99 x 109 N m2/C2), Q is the charge of the sphere, and r is the distance from the center of the sphere.
Treat the 3D sphere as a 2D circle. The radius for the sphere is the same radius as for the circle. No matter where on the sphere you place a mark, the distance (radius) from the mark to the centre of the sphere will always be the same as the circle.