011 10.0 points Two conducting spheres with diameters of 0.44 m and 0.7 m are se
ID: 3308498 • Letter: 0
Question
011 10.0 points Two conducting spheres with diameters of 0.44 m and 0.7 m are separated by a distance that is large compared to the diameters. The spheres are connected by a thin wire and are charged to 6 . What is the potential of the system of spheres when the reference potential is taken to be 0 at oo? The permittivity of a vacuum is 8.8542 × 10-12 C2/N·m2 . Answer in units of kV 012 10.0 point:s The electric potential in a certain region is 6.2 V/m. = where = 9 V/m*, 5.5 m2, and 7.7 V/m4. What is the y component of the electric field E,, at (5.1 m,-4 m. 5.3 m)? Answer in units of V/m 013 10.0 points Consider an equilateral triangle with sides of lengths 3 m and charge-0.2 C, 1.7 C and 1.3 C located at the corners of the triangle. Find the minimum work required to move the first point charge to infin- The permittivity of free space is it 8.85419 × 10-12 C2/N·m2 Answer in units of J.Explanation / Answer
011:
potential on surface of both spheres will be same.
V = k Q1 / r1 = k Q2 / r2
Q1 = Q2 (0.44/0.7)
and Q1 + Q2 = 6 + 6
(0.44/0.7) Q2 + Q2 = 12
Q2 = 7.368 uC
V = ( 9 x 10^9) (7.368 x 10^-6) / (0.7)
V = 94.7 x 10^3 Volt
Ans: 94.7 kV
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Ey = d(V) / dy
= 2 alpha x^2 y + 3 & y^2 z
= (2 x 9 x 5.1^2 x -4) + (3 x 7.7 x (-4)^2 x 5.3)
= 86.16 V/m .....Ans
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Potential at the location of first point,
V = k Q1 / r + k Q2 / r
V = (9 x 10^9 / 3 x 10^-6) (1.7 + 1.3)(10^-6)
V = 9 x 10^9 Volt
Work = q delta V
= (0.2 x 10^-6)(9 x 10^9)
= 1800 J
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