Two thin plastic spherical shells (shown in cross section in the figure below) a

Question

Two thin plastic spherical shells (shown in cross section in the figure below) are uniformly charged. The center of the larger sphere is at (0, 0); it has a radius of 12 cm and a uniform positive charge of + 2 times 10^-9 C. The center of the smaller sphere is at (26 cm, 0); it has a radius of 3 cm and a uniform negative charge of -9 times 10^-9 C. What are the components E_A, X and E_A, Y of the electric field E^rightarrow at location A (6 cm to the right of the center of the large sphere)? Neglect the small contribution of the polarized molecules in the plastic, because the shells are very thin and don't contain much matter. (Indicate the direction with the sign of your answer.) E_A, X = 2025 N/C E_A, Y = 0 N/C What are the components E_B, X and E_B, Y of the electric field at location B (15 cm above the center of the small sphere)? Neglect the small contribution of the polarized molecules in the plastic, because the shells are very thin and don't contain much matter. (Indicate the direction with the sign of your answer.) E_B, X = 173.3604679 N/C E_B, Y = -2721.126491 N/C What are the components F_X and F_Y of the force on an electron placed at location B? (Indicate the direction with the sign of your answer.) F_X = _____ N F_Y = _____ N

Explanation / Answer

Given
   plastic charged spheres of radii r1= 12 cm, r2 = 3 cm

   placed on the x axis at (0,0) and (26,0) cm

charges of q1 = 2*10^-9 C, q2 = -9*10^-9 C

     
the electric field at point B which is at (15,15)cm ,

   so it would be at distance from centre of A is r1 = sqrt(15^2+15^2) cm= 21.21 cm

due to q1 the field at B along y direction is By1 = k*q1/r1^2 = 9*10^9*2*10^-9/(0.2121)^2 N = 400 N/C

due to charge q2 field at B along y dierection is downward By2 = k*q1/r1^2 = 9*10^9*9*10^-9/(0.15)^2 N = 3600 N/C

now the field at B is By = -3600+400 = -3200 N/C

c) the force at point B is

   x component due to q2 is zero because the force is along y direction , only y component will be there

   due to charge q1 the force is F1 = kq1*q/r1^2

                   = 9*10^9*2*10^-9*1.6*10^-19/(0.2121)^2 N
                  
                   = 6.40193*10^-17 N
                      
   now F1x = F1 cos 225 = 6.40193*10^-17*cos225 N = (-4.526848)*10^-17 N
  
   F1y = F1 sin 225 = 6.40193*10^-17*sin225 N = (-4.526848)*10^-17 N

the F2y = kq2*q/r2^2 = 9*10^9*9*10^-9*1.6*10^-19/(0.15)^2 N = 5.76*10^-16 N

net Fy = 5.76*10^-16 -4.526848*10^-17 N = 5.3073152*10^-16 N

Fx = (-4.526848)*10^-17 N

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