Two thin plastic spherical shells are uniformally charged. Thecenter of the larg

Question

Two thin plastic spherical shells are uniformally charged. Thecenter of the larger sphere is at (0,0); it has a radiusof   12 cm and a uniform positive charge of +4*10^-9 C. The center of the smaller sphere is at(25 cm,0); it hasa radius of 3 cm and a uniform negative charge of -10^-9 C. A) what are the components Ea,x and Ea,y of the electric fieldchange of E at location "a" ( 6 cm to the right of the center ofthe large sphere)? Neglect the small contribution of the polarizemolecules in the plastic, because the shells are very thin anddon't contain much matter.    B) what are the components Eb,x and Eb,y of the electric fieldat location B (15 cm above the center of the small sphere)? Neglectthe small contribution of the polarized molecules in the plastic,because the shells are very thin and don't contain muchmatter. C) what are components Fx and Fy of the force on an electronplaced at location B? Two thin plastic spherical shells are uniformally charged. Thecenter of the larger sphere is at (0,0); it has a radiusof   12 cm and a uniform positive charge of +4*10^-9 C. The center of the smaller sphere is at(25 cm,0); it hasa radius of 3 cm and a uniform negative charge of -10^-9 C. A) what are the components Ea,x and Ea,y of the electric fieldchange of E at location "a" ( 6 cm to the right of the center ofthe large sphere)? Neglect the small contribution of the polarizemolecules in the plastic, because the shells are very thin anddon't contain much matter.    B) what are the components Eb,x and Eb,y of the electric fieldat location B (15 cm above the center of the small sphere)? Neglectthe small contribution of the polarized molecules in the plastic,because the shells are very thin and don't contain muchmatter. C) what are components Fx and Fy of the force on an electronplaced at location B?

Explanation / Answer

A)At location a,the elecric field components Eaxand Eay due to the two shells is Eax = E1x + E2x =E1 * cos1 + E2 *cos2 --------(1) and Eay = E1y + E2y =E1 * sin1 + E2 *sin2 ----------(2) E1 = k *(q1/r12) k = (1/4o) = 9 * 109Nm2/C2,q1 = +4 * 10-9 Cand r1 = 12 + 6 = 18 cm = 18 * 10-2 m and E2 = k *(q2/r22) q2 = -1 * 10-9 C and r2= 25 - 6 = 19 cm = 19 * 10-2 m the angles are 1 = 2 =0o from (1) and (2) we get Eax = E1 * cos1 +E2 * cos2 = E1 *cos(0o) + E2 * cos(0o) =E1 + E2 In the second equation,the sine components become zerotherefore we get Eay = 0 B)At location B,the elecric field components Ebxand Eby due to the two shells is Ebx = E1x + E2x =E1 * cos1 + E2 *cos2 --------(1) and Eby = E1y + E2y =E1 * sin1 + E2 *sin2 ----------(2) the angles are tan1 = (15 - 0/0 - 12) = (-15/12) or 1 = tan-1(-1.25) =-51.3o and 2 = 90o C)At location B,the force components Fx andFy due to the two shells is Fx = F1x + F2x =F1 * cos1 + F2 *cos2 and Fy = F1y + F2y =F1 * sin1 + F2 *sin2 F1 = k * (q1 *e/a12) e = -1.6 * 10-19 C and a12 =((0 - 12)2 + (15 - 0)2) = 369 cm2= 369 * (10-2)2 m2 = 369 *10-4 m2 and F2 = k * (q2 *e/a22) and F2 = k * (q2 *e/a22) a22 = ((0 - 0)2 + (15 -3)2) = 144 cm2 = 144 *(10-2)2 m2 = 144 *10-4 m2 the angles are 1 = -51.3o and2 = 90o

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