Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by
where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so
And so, the work done is also zero.
Given :
An electron moving in the positive x direction experiences a magnetic force in the positive z direction.
To Find :
The direction of the magnetic field.
Solution :
We know, force is given by :
Here, q = -e.
Now, for above condition to satisfy :
So, 
Therefore, direction of magnetic field is negative y direction.
Hence, this is the required solution.
Answer:
40 V
Explanation:
I will assume that the resistors are
100 and 3900 and 1000 OHMS <=====(NOT W)
In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total
V = IR
I = V / R so the total current will be 200 v / 5000 ohms = .04 amps
this is the current through all of the resistors
so for the 1000 ohm resistor V = IR .04 (1000) = 40 V
I would say your answer is B- Some of the chemical energy from the batteries is converted into heat energy.
Answer:
64 J
Explanation:
The potential energy change of the spring ∆U = -W where W = work done by force, F.
Now W = ∫F.dx
So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)
∆U = - ∫-Fdx
= ∫F.dx
Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx
= ∫₀²(40x - 6x²).dx
= ∫₀²(40xdx - 6x²dx)
= ∫₀²(40x²/2 - 6x³/3)
= ∫₀²(20x² - 2x³)
= [20x² - 2x³]₀²
= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)
= [(20(4) - 2(8)) - (0 - 0))
= [80 - 16 - 0]
= 64 J
