Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
Answer:
Moving a unit "positive" test charge from A to B will result in a reduction in potential
V = K Q / R potential at a point
V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q
V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8
V2 - V1 = -4.17 * 562.5 J/C
V = - 2346 Volts
North: 1m
South: 0,8m
Direction:
1m>0,8m
so mouse moved north
Distance:
1m-0,8m=0,2m
so mouse traveled 0,2m
Answer: The mouse moved 0,2m north.
"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."
Regards M.Y.