Question

Example 3.2. A point charge q is situated a distance a from the center of a grounded conducting sphere of radius R (Fig. 3.12). Find the potential outside the sphere. b q' V = 0 FIGURE 3.12 FIGURE 3.13

Answer 1

bq! V-0 a he q' in the image charge at a distance b from the center of ophere. 6LR. change a in inside the sphere. from fig. 3. 13 we get r = fo2ta ²_zracoso and r' = 62+6² - arbcoso fig-3.13. at in convenient to formulate the problem by means of spherical Corordinates with the origin of Coordinater at the center of sphere. The boundary condition p (r=R) = 0 can be satisfied by an image charge q inside the sphere. bqla 9

At the surface potential vao. At point of the potential in Arcto a R + 1 to 2 0 а - С R At the point of the potential in the ho atr + te h o >> a r ed - from ① and ② we get after rearranging 9/R- b) = a (a-R) qR-qb= q R-g'a 3 and a (Rab) = -9' (QAR) 2) a Rtab = - g'a- a'r - Adding and we get 29R =-2g'a = a = - 9R

subtracting (4) from 3 we get - 296 = +29'k = b = - B.al = - B. (-9R) => b= R2 - Unitzber 2016 (656 246² - 2nbCoso & J 22 +64 - 20 R2 coso ale raa ( m2 + R - 20 R2 caso) -9 J (ar) 2 + R2 _ eracuso - The potential at A due to q and a V = areto (a + a in

watch V on to ara uso 416 2a²- 2ra coso 12 - 2ra coso rr2a²-aracoso √ R² + (raja-aracoso This in the potential outside Conducting the grounded