An electric flux of 147 n*m^2/c passes through a flat horizontal surface that has an area of 0.824 m^2. the flux is due to a uniform electric field. what is the magnitude of the electric field if the field points 31.6° above the horizontal?

231.16 N/C

Explanation:

Given

Electric Flux

Area(A)

Given Field point above

Therefore angle between Area vector Electric Field =90-31.6=

We know that Flux is given by

E=231.16 N/C

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j