A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
1 answer:
Answer: Option (b) is the correct answer.
Explanation:
It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.
Then test charge to the right immediately after being released.
Therefore, the net force will be as follows.
F =
=
=
F = > 0
Thus, we can conclude that the test charge move to the right immediately after being released.
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Answer:
K.E = 100 J
Final P.E = 100 J
Explanation:
The kinetic energy of any object can be given by the following formula:
where,
K.E = Kinetic Energy
m = mass of ball = 2 kg
v = speed of ball
Initially, v = 10 m/s. Therefore, the initial K.E is given as:
<u>K.E = 100 J</u>
Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:
Initial K.E + Initial P.E = Final K.E + Final P.E
Initial P.E is also zero due to zero height initially.
K.E + 0 = 0 + Final P.E
<u>Final P.E = 100 J</u>