A sphere with radius 70 cm has its center at the origin.Equal charges of 3 C are

Question

A sphere with radius 70 cm has its center at the origin.Equal charges of 3 C are placed at 72 degree intervals alongthe equator of the sphere. The Coulomb constant is 8.99E9 Nm^2/C^2. a) What is the electric potential at the origin? Answer inunits of kV. b) What is the electric potential at the north pole? Answer inunits of kV. A sphere with radius 70 cm has its center at the origin.Equal charges of 3 C are placed at 72 degree intervals alongthe equator of the sphere. The Coulomb constant is 8.99E9 Nm^2/C^2. a) What is the electric potential at the origin? Answer inunits of kV. b) What is the electric potential at the north pole? Answer inunits of kV.

Explanation / Answer

a)The electric field inside a sphere is E = ke* (Q/a3) * r ke= (1/4o) = 9 *109 Nm2/C2,Q is the charge on thesphere,a is the radius of the sphere and r is thedistance within the sphere from the center of thesphere. at the origin r = 0 therefore we get E =  ke* (Q/a3) * 0 = 0 therefore,the electric potential at the origin is E = V * r or V = (E/r) = (0/r) = 0 b)the net electric field at the north pole is E = (Ex2 +Ey2)1/2 -----------(1) Ex = E1x + E2x =E1 * cos1 + E2 *cos2 and Ey = E1y + E2y =E1 * sin1 + E2 *sin2 E1 = ke*(q1/r12) q1 = 3 C = 3 * 10-6 C andr1 = a * 1 = a * (/2) = a *(3.14/2) = 1.57a a = 70 cm = 70 * 10-2 m or r1 = 1.57 * 70 * 10-2 = 1.099 m and E2 = ke*(q2/r22) q2 = 3 C = 3 * 10-6 C andr2 = a * 2 = a * (90o -72o) = a * 18o = a * (/10)= a * (3.14/10) = 0.314a or r2 = 0.314 * 70 * 10-2 = 0.2198m therefore,the electric potential at the north pole is V = E * a the value of E is obtained from equation (1) Ex = E1x + E2x =E1 * cos1 + E2 *cos2 and Ey = E1y + E2y =E1 * sin1 + E2 *sin2 E1 = ke*(q1/r12) q1 = 3 C = 3 * 10-6 C andr1 = a * 1 = a * (/2) = a *(3.14/2) = 1.57a a = 70 cm = 70 * 10-2 m or r1 = 1.57 * 70 * 10-2 = 1.099 m and E2 = ke*(q2/r22) q2 = 3 C = 3 * 10-6 C andr2 = a * 2 = a * (90o -72o) = a * 18o = a * (/10)= a * (3.14/10) = 0.314a or r2 = 0.314 * 70 * 10-2 = 0.2198m therefore,the electric potential at the north pole is V = E * a the value of E is obtained from equation (1)

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