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.
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V

Therefore, the resistance that will provide this potential drop is 388.89 ohms.
Answer:
longitudinal and transverse.
Explanation:
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