An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of th
e jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?
A) We’ll see the bell move, but we won’t hear it ring.
B) We won’t see the bell move, but we’ll hear it ring.
C) We’ll see the bell move and hear it ring.
D) We won’t see the bell move or hear it ring.
E) We’ll see the sound waves exit the vacuum pump.
A) We’ll see the bell move, but we won’t hear it ring.
B) We won’t see the bell move, but we’ll hear it ring.
C) We’ll see the bell move and hear it ring.
D) We won’t see the bell move or hear it ring.
E) We’ll see the sound waves exit the vacuum pump.
2 answers:
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Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
"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.