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State Gauss's theoremin electrostatics.Apply this theorem to derive an expression for electric field intensityat a point outside a uniformlycharged thinsphericalshell. |
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Answer» Solution :Gauss's Theorem in electrostatics : It states that the flux of electric field through any closed surface `delta` in the vacuum is `(1)/(epsilon_(0))` TIMES the total charges enclosed by S. `:. phi = oint vec(E).dvec(s)=(q)/(epsilon_(0))` Electric field due to a uniform charged spherical shell : Suppose a thin spherical shell of Radius R and centre O. Let the charge +q is distributed over the surface of sphere. Electric field intensity `vec(E)` is same at every POINT on the surface of sphere directly outward. Let a point P outside the shell with radius vector `vec(r)` and small area element `vec(dS) = hat(n) dS`. According to Gauss's law `oint vec(E).vec(dS)=(q)/(epsilon_(0))RARR oint E dS = (q)/(epsilon_(0))` Since `vec(E) and hat(n) ` arein the same direction `E 4 pi r^(2) = (q)/(epsilon_(0))rArr E = (1)/(4 pi epsilon_(0)) (q)/(r^(2))` Vectorially,`vec(E)=(1)/(4 pi epsilon_(0))(q)/(r_(0))hat(r)` Special cases (i) At the point on the surface of the shell, `r=R " " :. " " E=(1)/(4pi epsilon_(0))(q)/(R^(2))` (ii) If `sigma` is the surface chargedensity on theshell then `q=4 pi r^(2) sigma` `:. E=(1)/(4pi epsilon_(0)) (4 pi R^(2)sigma)/(R^(2))=(sigma)/(epsilon_(0))` (iii) If the point P lies INSIDE hte spherical shell then the Gaussian surface encloses no charge `:. q=0` Hence E=0 `##SB_PHY_XII_08_DB_E01_025_S01.png" width="80%"> |
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