Refer to the figure below.
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
Water was confirmed to be on the sunlit surface of the Moon
Answer: The correct answers are (A) and (C).
Explanation:
The expression from electrostatic force is as follows;

Here, F is the electrostatic force, k is constant, r is the distance between the charges and
are the charges.
The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.
Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.
The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.
Therefore, the true statements from the given options are as follows;
Like charges repel.
Unlike charges attract.
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.