Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:
In our problem we have 
and 
, so we can find the magnitude of the electric field:
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
Therefore, the magnitude of the force acting on the proton will be
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
Answer:
Not possible
Explanation:
Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.
So before the collision, 1 particle is at rest, 1 not -> total momentum is non-zero
After the collision, both particles are at rest -> total momentum is zero which is different from before.
Therefore this is not possible.
Answer:
Explanation:
For the first case , the expression for electrostatic force can be given by the following .
F = K x 8Q x 2Q / r² where k is a constant .
F = K 16 Q² / r²
When they touch , some charge is neutralized . Net charge remaining
= 8Q - 2 Q = 6 Q
Charge on each sphere = 6Q/2 = 3 Q .
Force between them
F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²
F₁ / F = 9 / 16
F₁ = 9 F / 16 .
