A solid enclosed between two concentric spheres is called a spherical shell. The table of contents will list only tasks having one of the required ranks in corresponding rankings and at least one of the required tags (overall). First, we need to recall just how spherical coordinates are defined. Where: Input Volume Data. Now we evaluate the charge Q1. We choose a surface of a sphere to be the Gaussian surface. Zero potential is selected in infinity. Therefore, the entire integral is divided into two parts. and evaluate the magnitude of the electric field inside the charged spherical shell. We factor the expression in parentheses. Because the volume of water that flows from the container is equal to the volume of the spherical object. Then we divide the second integral into two integrals: and calculate the integrals. These two equations show that the electric potential is continuous at … Inside the Gaussian surface there is the whole charged shell, thus the charge can be evaluated through the shell volume V and the charge density ρ. 2) Determine also the potential in the distance z. The sphere will displace a volume of water equal to that of itself, and this shows how much the water will rise. Outside Radius R (in, mm) =. First we have to transfer the charge from infinity to the surface of the shell (i.e. The potential at a given point is equal minus the intensity integrated from the place of zero potential to the given point. Giving the fact that it is useful to apply Gauss's law to solve this problem, it is necessary to decide what to choose for the Gaussian surface. Therefore, the volume of the spherical shell formed between the concentric circles would be the difference between the outer sphere and the inner sphere--which, when mathematically written, would be 4 3 π R 3 − 4 3 π r 3. Now it remains to evaluate the potential inside the sphere. it is smooth. We substitute the magnitude of the electric intensity in the integral. The pore volume corresponds to what we call the free volume, while the spherical volume of radius (R−D/2) is called the cavity, where D is the particle diameter. The volume of charges in the shell of infinitesimal width is equal to the product of the area of surface and the thickness . Giving the fact, that the electric field intensity depends only on the distance from the centre of the shell, the potential is also dependent only on the distance z from the centre of the shell. If we reduce the hollow part of the charged shell until a = 0 and denote b = R, we obtain a ball of radius R charged with charge density ρ. So we're 60% of the way. We examine separately the field outside the spherical shell, inside the shell and inside the hollow part. (A more detailed explanation is given in the section Hint.). When calculating the potential inside the sphere we proceed as in the previous section. Since this selected area is within the charged shell, no charge is enclosed in it, i.e Q = 0. This expression can be used to calculate the exact volume of a sphere composed of a small number of shells with finite thickness $h$. In the -dimensional velocity subspace of the -dimensional phase space the condition again defines a spherical shell whose volume is essentially equal to the volume of the enclosed sphere. The charge closed in the surface is also given by the charge density and the volume of the part of the shell that is closed in the Gaussian surface. Find the radius of the sphere which is formed. The magnitude of electric field intensity inside the shell (a < z < b) is, The magnitude of electric field intensity outside the charged spherical shell (z > b) is. Your email address will not be published. Force F divided by charge Q is electric field intensity \(\vec{E}\). If the sphere is completely submerged, how much will the water rise? The potential therefore remains constant and of the same value as on the inner surface of the shell. The expression in the parentheses is factored as in the case of electric intensity. When calculating the potential inside the charged spherical shell and within the hollow part, we must be careful, because the electric field intensity is not given by the same relation along the path of integration; it is described by different equations outside and inside of the shell and inside the hollow part. $$\therefore $$  Volume of sphere $$ = \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {\left( {\frac{5}{2}} \right)^3} = \frac{{125}}{6}\pi $$cu.cm, The radius of the cylindrical vessel $$ = \frac{{10}}{2} = 5$$cm, The volume height occupied by $$\frac{{125}}{6}\pi $$cu.cm. For a spherical shell, if $$R$$ and $$r$$ are the outer and inner radii respectively, then the volume of the shell is, $$ = \frac{4}{3}\pi \left( {{R^3} – {r^3}} \right)$$, $$ = \frac{\pi }{6}\left( {{D^3} – {d^3}} \right)$$. Thus, the charge is evaluated using the charge density and a volume of the sphere part enclosed in the Gaussian sphere, We substitute both gained relations into Gauss's law (*). The radius of the Gaussian sphere is smaller than the inner radius of the charged spherical shell. A sphere of radius 5cm is dropped into a cylindrical vessel partly filled with water. The wall thickness must be small (r/t> 10) 2. find the behaviour of the electric intensity and the electric potential depending on the variable z in the interval "from zero to infinity". The first parenthesis in the numerator is equal to the thickness of the sphere ΔR. When calculating the potential inside the hollow part we proceed as in the previous section. We have obtained a relation for evaluating the electric field intensity outside the charged sphere. Electric potential at point A is equal to a negative taken integral of intensity from a point of zero potential to the point A. The relation for intensity changes on the surface of the shell. So let's get started. The function is continuous. A nonconducting spherical shell of inner radius R 1 and outer radius R 2 contains a uniform volume charge density ρ throughout the shell. Solution: Step 1: The charge distribution is spherically symmetric. At a distance z the electric field intensity of the charged shell is: In this section we evaluate the intensity of the electric field inside the charged spherical shell (a < z < b ). The vector of electric field intensity \(\vec{E}\) is parallel to the vector \(\vec{z}\). The integral is therefore equal to. 0 0. Volume of a spherical shell? We multiply the parentheses and add the same components together. The Volume of the Sphere = 4 / 3 πr 3 The Volume of Hemisphere As already said a hemisphere is half the sphere, hence its volume will also be half the volume of the sphere. Multiplying the volume with the density at this location, which is , gives the charge in the shell: Only a portion of the charge is closed in the Gaussian surface. In this task for an integration curve we choose a part of a line leading through the midpoint of the sphere. We choose the Gaussian surface to be a surface of a sphere, which is centred in the midpoint of the charged sphere and its radius is z, a < z < b. A spherical shell or hollow sphere is made of two spheres of different sizes and with the same center, where the smaller sphere is subtracted from the larger. Result. Spherical Shell Calculator. Therefore, we set up the problem for charges in one spherical shell, say between and , as shown in Figure 2.3.6. A hollow sphere is a ball that has been hollowed such the an equal thickness wall creates anopther internal ball within the external ball. The intensity increases over the interval a to b and then in the distance z greater than bthe intensity decreases with the square of the distance z. Stresses that act tangentially to the curved surface of a shell are known as membrane stresses. The radius of the Gaussian sphere is larger than the outer radius of the charged spherical shell. We substitute the limits of the integral and obtain the size of the potential outside the shell at distance z: Note: If we substitute the total charge \(Q\,=\, \frac{4}{3}\pi \left(b^3-a^3\right)\varrho\) in the equation, we obtain the same relation as the relation for the potential around a point charge. If we enlarge the hollow part inside the shell until a = b = R, we obtain a sphere of radius R. Instead of charge bulk density, we consider area density. This time the electric field intensity is not represented by the same relation along the integral curve. Thus we obtain the equation for potential inside the charged shell. If P is an external point, in order to find the field due to the entire spherical shell, we integrate from ξ = r − a to r + a. Section 4-7 : Triple Integrals in Spherical Coordinates. We divide this task into three parts. The field around a charged spherical shell is therefore the same as the field around a point charge. \], \[\oint_c \vec{E} \cdot \mathrm{d}\vec{S}\,=\, \frac{Q_1}{\varepsilon_0}\,,\tag{*}\], \[\oint_c \vec{E} \cdot \mathrm{d}\vec{S}\,=\,\oint_c En \mathrm{d}S\,=\,\oint_c E \mathrm{d}S\,=\,E\oint_c \mathrm{d}S\,,\], \[\oint_c \vec{E} \cdot \mathrm{d}\vec{S}\,=\,E\, 4 \pi z^2\,\], \[Q_1\,=\,V_1 \varrho \,=\, \frac{4}{3} \pi \left(z^3-a^3\right) \varrho\,.\], \[E \,4\pi z^2\,=\, \frac{\frac{4}{3} \pi \left(z^3-a^3\right)\varrho}{\varepsilon_0}\], \[E \,=\, \frac{\varrho}{3 \varepsilon_0}\,\frac{z^3-a^3}{z^2} \,=\, \frac{\varrho}{3 \varepsilon_0}\,\left(z-\frac{a^3}{z^2} \right) \], \[\oint_S E \cdot \mathrm{d}S \,=\, \frac{Q}{\epsilon_0}\,,\], \[\varphi (z)\,=\, - \int_{\infty}^z \vec{E} \cdot \mathrm{d}\vec{z}\], \[ \varphi (z)\,=\, - \int^{z}_{\infty} E \mathrm{d}z \], \[\varphi (z)\,=\, - \int^{z}_{\infty} E \mathrm{d}z \], \[E \,=\, \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\frac{1}{z^2}\], \[\varphi (z)\,=\, - \int^{z}_{\infty} \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\frac{1}{z^2} \,\mathrm{d}z \,=\, - \,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0} \int^{z}_{\infty} \frac{1}{z^2}\, \mathrm{d}z\], \[\varphi (z)\,=\,- \,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\left[- \frac{1}{z}\right]^z_{\infty}\,.\], \[\varphi (z)\,=\,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\frac{1}{z}\,.\], \[ \varphi (z)\,=\, \frac{Q}{4 \pi \varepsilon_0}\,\frac{1}{z}\], \[ \varphi (z)\,=\, - \int^{z}_{\infty} \vec{E} \cdot \mathrm{d}\vec{z}\,=\, - \int^{z}_{\infty} E \mathrm{d}z \,.\], \[\varphi (z)\,=\, - \int^{b}_{\infty} E_v \mathrm{d}z - \int^{z}_{b} E_u \mathrm{d}z \], \[\varphi (z)\,=\, - \int^{b}_{\infty} \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\, \frac{1}{z^2}\, \mathrm{d}z \, - \int^{z}_{b} \frac{ \varrho}{3 \varepsilon_0\,}\, \left(z-\frac{a^3}{z^2}\right) \, \mathrm{d}z \,.\], \[\varphi (z)\,=\, - \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\int^{b}_{\infty} \frac{1}{z^2}\, \mathrm{d}z \,- \,\frac{ \varrho}{3 \varepsilon_0\,}\int^{z}_{b} \left(z-\frac{a^3}{z^2}\right)\, \mathrm{d}z \,,\], \[\varphi (z)\,=\, - \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\int^{b}_{\infty} \frac{1}{z^2} \,\mathrm{d}z \,-\, \frac{ \varrho}{3 \varepsilon_0\,}\int^{z}_{b} z\, \mathrm{d}z \, + \,\frac{ \varrho}{3 \varepsilon_0\,}\int^{z}_{b} \frac{a^3}{z^2}\, \mathrm{d}z\,,\], \[\varphi (z)\,=\, - \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\left[- \frac{1}{z}\right]^b_{\infty}- \,\frac{ \varrho}{3 \varepsilon_0\,}\left[\frac{z^2}{2}\right]^z_{b} + \,\frac{ \varrho}{3 \varepsilon_0\,}\left[-\frac{a^3}{z}\right]^z_{b}\,.\], \[\varphi (z)\,=\,- \frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\left(- \frac{1}{b}\right)\,- \,\frac{ \varrho}{3 \varepsilon_0\,}\left(\frac{z^2}{2}-\frac{b^2}{2}\right) \,+ \,\frac{ \varrho}{3 \varepsilon_0\,}\left( -\frac{a^3}{z}\,+\frac{a^3}{b}\right)\,.\], \[\varphi (z)\,=\,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0} \frac{1}{b}\,- \,\frac{ \varrho}{3 \varepsilon_0}\left(\frac{z^2}{2}-\,\frac{b^2}{2}\right) \,+ \,\frac{ \varrho}{3 \varepsilon_0\,}\left( -\frac{a^3}{z}\,+\frac{a^3}{b}\right)\], \[\varphi (z)\,=\, \frac{\varrho b^3}{3 \varepsilon_0\, b} \,-\,\frac{\varrho a^3}{3 \varepsilon_0 b} \,-\,\frac{ \varrho z^2}{6 \varepsilon_0} \,+\,\frac{ \varrho b^2}{6 \varepsilon_0} \,- \,\frac{ \varrho a^3}{3 \varepsilon_0 \,z } \,+ \,\frac{ \varrho a^3}{3 \varepsilon_0\, b}\], \[\varphi (z)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho z^2}{6 \varepsilon_0} \,- \,\frac{ \varrho a^3}{3 \varepsilon_0 \,z } \], \[\varphi (z)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \left( \frac{ z^2}{2 } +\,\frac{ a^3}{z}\right) \], \[\varphi (z)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \left( \frac{ z^2}{2 }+\,\frac{ a^3}{z}\right) \,=\,\frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \left( \frac{ a^2}{2 } +\,\frac{ a^3}{a}\right)\], \[ \varphi (z)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \left( \frac{ a^2}{2 } +\, a^2 \right)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \frac{ 3a^2}{2 }\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho a^2}{2 \varepsilon_0} \], \[ \varphi (z)\,=\,\frac{\varrho}{2 \varepsilon_0} \left(b^2-a^2\right)\], \[E \,=\, \frac{\varrho \left(b^3 - a^3\right)}{3 \varepsilon_0} \, \frac{1}{z^2}\,=\, \frac{Q}{4 \pi \varepsilon_0} \, \frac{1}{z^2} \,.\], \[E \,=\, \frac{\varrho}{3 \varepsilon_0}\,\left(z-\frac{a^3}{z^2} \right) \,.\], \[\varphi (z)\,=\,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\frac{1}{z}\,=\, \frac{Q}{4 \pi \varepsilon_0} \, \frac{1}{z}\,.\], \[\varphi (z)\,=\, \frac{\varrho b^2}{2 \varepsilon_0} \,-\,\frac{ \varrho }{3 \varepsilon_0} \left( \frac{ z^2}{2 }+\,\frac{ a^3}{z}\right)\,. We can evaluate the electric field intensity of a charged ball by using the above derived results. The potential does not depend on the choice of the integration curve; therefore it can be freely selected. Volume of the cylinder $$ = \pi {r^2}h = \pi \times {\left( 9 \right)^2} \times 12$$ Consider a spherical shell of inner radius ri and outer radius ro. The surface to volume ratio of this spherical segment = 1.24 Surface area to volume ratio is also known as surface to volume ratio and denoted as sa÷vol, where sa is the surface area and vol is the volume. (1) = (2) $$ \Rightarrow \frac{4}{3}\pi {r^3} = 972\pi $$ The wall of a pressurized spherical vessel is subjected to uniform tensile stresses σin all directions. This is due to the symmetrical distribution of a positive charge in the shell (If the charge were negative, the vectors would be of an opposite direction). We choose the Gaussian surface to be a surface of a sphere with a radius z, its midpoint being the midpoint of the charged shell. When computing the intensity inside the hollow part of the spherical shell, the radius of the Gaussian sphere is smaller than the inner radius of the shell. \[E_p(z)\,=\, - \int^z_{\infty} \vec{F} \cdot \mathrm{d}\vec{z}\], \[\varphi\,=\, - \int^z_{\infty} \frac{\vec{F}} {Q}\cdot \mathrm{d}\vec{z}\], \[\varphi\,=\, - \int^z_{\infty} \vec{E}\cdot \mathrm{d}\vec{z}\], \[\oint_S \vec{E} \cdot \mathrm{d}\vec{S}\,=\, \frac{Q}{\varepsilon_0}\,.\], \[\oint_S \vec{E} \cdot \vec{n}\mathrm{d}S\,=\, \frac{Q}{\varepsilon_0}\tag{*}\], \[\oint_c \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,\oint_c E n\mathrm{d}S\,=\, \oint_c E\mathrm{d}S\,.\], \[\oint_c \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,E \oint_c \mathrm{d}S\,.\], \[\oint_c \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,E S_s\,,\], \[\oint_c \vec{E} \cdot \vec{n}\mathrm{d}S\,=\,E\, 4 \pi z^2\], \[E 4 \pi z^2\,=\, \frac{Q}{\varepsilon_0}\], \[E \,=\, \frac{1}{4 \pi \varepsilon_0}\,\frac{Q}{z^2}\tag{**}\], \[Q\,=\,V \varrho\,=\,\frac{4}{3} \pi \left(b^3 - a^3\right) \varrho\], \[E \,=\, \frac{1}{4 \pi \varepsilon_0}\,\frac{\frac{4}{3} \pi \left(b^3 - a^3\right) \varrho}{z^2}\,=\, \frac{\varrho \left(b^3 - a^3\right)}{3 \varepsilon_0\,z^2}\], \[E \,=\, \frac{\varrho \left(b^3 - a^3\right)}{3 \varepsilon_0} \, \frac{1}{z^2}\,. Multiplying the volume with the density at this location, which is, gives the charge in the shell: Spherical symmetry with non-uniform charge distribution. Using this knowledge we can adjust the integral on the left side of the Gauss's law: The vector of electric field intensity E is at all points of the Gaussian surface of the same magnitude, so we can factor the electric intensity out of the integral as a constant. Graph of the electric potential as a function of the distance from the centre of the shell: The electric potential inside the charged spherical shell is equal to. The procedure is very similar to the previous section: The intensity outside the charged shell, therefore this solution is not described in detail. Inside it the whole charge is distributed in the shell. We factor all constants out of the integral. The intensity of the electric field inside the hollow part of the charged shell (z < a) can be determined again by using Gauss's law. When calculating the intensity field in the charged shell, the radius of the Gaussian sphere is smaller than the outer radius of the shell and greater than the inner radius of the shell. \], \[E \,=\, \frac{\varrho \left(b^3 - a^3\right)}{3 \varepsilon_0} \, \frac{1}{z^2} \,=\, \frac{\varrho \left(b - a\right) \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0} \, \frac{1}{z^2} \], \[E \,=\, \frac{\varrho \mathrm{\Delta}R \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0} \, \frac{1}{z^2}\,=\, \frac{\sigma \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0} \, \frac{1}{z^2} \], \[E \,=\, \frac{\sigma \left(R^2 +R^2+ R^2\right)}{3 \varepsilon_0} \, \frac{1}{z^2}\,=\, \frac{\sigma 3R^2 }{3 \varepsilon_0} \, \frac{1}{z^2} \,=\, \frac{\sigma R^2 }{ \varepsilon_0} \, \frac{1}{z^2}\], \[\varphi (z)\,=\,\frac{\varrho \left(b^3-a^3\right)}{3 \varepsilon_0}\,\frac{1}{z}\,=\,\frac{\varrho \left(b - a\right) \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0}\frac{1}{z}\], \[\varphi (z)\,=\, \frac{\varrho \mathrm{\Delta}R \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0} \, \frac{1}{z}\,=\, \frac{\sigma \left(b^2 +ba+ a^2\right)}{3 \varepsilon_0} \, \frac{1}{z}\,.\], \[\varphi (z)\,=\, \frac{\sigma \left(R^2 +R^2+ R^2\right)}{3 \varepsilon_0} \, \frac{1}{z}\,=\, \frac{\sigma 3 R^2}{3 \varepsilon_0} \, \frac{1}{z}\,.\], \[\varphi (z)\,=\, \frac{\sigma R^2}{ \varepsilon_0} \, \frac{1}{z}\,.\], \[ \varphi (z)\,=\,\frac{\varrho \left(b^2-a^2\right)}{2 \varepsilon_0} \,=\,\frac{\varrho \left(b-a\right)\left(b+a\right)}{2 \varepsilon_0} \,=\,\frac{\varrho \mathrm{\Delta}R \left(b+a\right)}{2 \varepsilon_0}\,=\,\frac{\sigma \left(b+a\right)}{2 \varepsilon_0}\,.\], \[ \varphi (z)\,=\,\frac{\sigma \left(R+R\right)}{2 \varepsilon_0}\,=\,\frac{\sigma 2R}{2 \varepsilon_0}\], \[ \varphi (z)\,=\,\frac{\sigma R}{ \varepsilon_0}\,.\], Tasks requiring comparison and contradistinction, Tasks requiring categorization and classification, Tasks to identify relationships between facts, Tasks requiring abstraction and generalization, Tasks requiring interpretation,explanation or justification, Tasks aiming at proving, and verification, Tasks requiring evaluation and assessment, Two balls on a thread immersed in benzene, Electric Intensity at a Vertex of a Triangle, A charged droplet between two charged plates, Capaciter partially filled with dielectric, Electrical Pendulum in Charged Sphere’s Field (Big Deflection), Gravitational and electric force acting on particles, Field of Charged Plane Solved in Many Ways, Electric resistance of a constantan and a copper wire, Electrical Resistances of Conductors of Different Lengths, Electrical Resistance of Wires of Different Cross Sections, Measuring of the electrical conductivity of sea water, Two-wire Cable between Electrical Wiring and Appliance, Using Kirchhoff’s laws to solve circiut with two power supplies, Change of the current through potentiometer, Application of Kirchhoff’s laws for calculation of total resistance in a circuit, Current-carrying wire in a magnetic field, Magnetic Force between Two Wires Carrying Current, Magnetic Field of a Straight Conductor Carrying a Current, Magnetic Field of a Straight Conductor inside a Solenoid, The motion of a charged particle in homogeneous perpendicular electric and magnetic fields, Voltage Induced in a Rotating Circular Loop, Inductance of a Coil Rotating in a Magnetic Field, The Length of the Discharge of the Neon Lamp, Instantaneous Voltage and Current Values in a Series RLC Circuit, RLC Circuit with Adjustable Capacitance of Capacitor, Heating Power of Alternating Current in Resistor, Resonance Frequency of Combined Series-Parallel Circuit.
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