A 92 cm diameter loop is rotated in a uniform electric field until the position of maximu electric flux is found. The flux in this position is measued to be 4.56 X 10 5 Nm^2/C what is the electirc field strength?

Answers should be in unit of N/C

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A 92 cm diameter loop is rotated in a uniform electric field until the position of maximu electric flux is found. The flux in this position is measued to be 4.56 X 10 5 Nm^2/C what is the electirc field strength?

Answers should be in unit of N/C

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## Answers (1)

look at the units of flux for a hint in solving this problem

flux = electric field x area x cos angle between field and a unit vector normal to the area

if the flux is maximum, then we know cos x =1, so that

flux = electric field x area

the area is pi r^2 = pi (0.46)^2 =0.66 m^2

therefore, E = flux/A = 4.56 x 10^5 Nm^2/C / 0.66m^2

=6.86x10^5 N/C