Draw equipotential lines near the positive and negative charges below with dashed lines. b) Draw solid electric field lines base
d on the fact that they are perpendicular to the equipotential lines at every point. Include arrows on the electric field lines to indicate their direction (since they represent vectors at each location).
1 answer:
Answer: find the attached figure for a and b
Explanation:
A) The second figure depict electric field lines and equipotential lines for two equal but opposite charges. The equipotential lines can be drawn by making them perpendicular to the electric field lines. The potential is greatest (most positive) near the positive charge and least (most negative) near the negative charge.
B) The figure attached depicts an isolated point charge Q with its electric field lines in blue and equipotential lines in green. The potential is the same along each equipotential line, meaning that no work is required to move a charge anywhere along one of those lines. Work is needed to move a charge from one equipotential line to another. Equipotential lines are perpendicular to electric field lines in every case.
Please find the attached file for the figure
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Answer:
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
Explanation:
Let the depth of each section is h.
That means the real depth for each section is h.
Apparent depth is liquid A is 7 cm.
Apparent depth in liquid B is 6 cm.
Apparent depth in liquid C is 5 cm.
by the formula of the refractive index
n = real depth / apparent depth
where, n is the refractive index of the liquid.
For liquid A:
.... (1)
For liquid B:
..... (2)
For liquid C:
..... (3)
By comparing all the three equations
nc > nB > nA
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
Answer:
From lowest to highest acceleration:
3rd train
2nd train
1st train
Explanation:
The acceleration of an object can be found by using Newton's second law:
where
a is the acceleration
F is the net force on the object
m is the mass of the object
We notice that for equal values of the forces F, the acceleration a is inversely proportional to the mass, m. Therefore, greater mass means lower acceleration, and viceversa.
So, the train with lowest acceleration is the one with largest mass, i.e. the 3rd train consisting of 50 equally loaded freight cars. Then, the 2nd train has larger acceleration, since it consists of 50 empty freight cars (so its mass is smaller). Finally, the 1st train (a single empty car) is the one with largest acceleration, since it is the train with smallest mass.