wo 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=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

vapour or gas assuming atmospheriv pressure

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west