Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.
Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface,
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :
So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
Answer:
Explanation:
In this case mechanical energy is conserved, which means that the sum of the initial kinetic energy and initial potential gravitational energy will be equal to the sum of the final kinetic energy and final potential gravitational energy:
Which in our case will be:
Which, since ,
,
,
and canceling <em>m</em> means that:
Solving for the final velocity we get: