Answer:
E=6v/d N/C
Explanation:
The question is "Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 1083 N/C. If the voltage is tripled and the distance between the plates is reduced to 1/2 the original distance, what is the magnitude of the new electric field"
E = V/d
where E= electric field
V= volage
d= separation distance
Given data
voltage is trippled= 3V
distance is reduced by 1/2 the original distance= d-d/2
= d/2
The magnitude of the new electric field is
E=3v/d/2
E=3v*2/d
E=6v/d N/C
