1- A uniform electric field of magnitude E = 250 N/C makes an angle ? = 45.0o wi

Question

1- A uniform electric field of magnitude E = 250 N/C makes an angle ? = 45.0o with the normal to the surface and the surface area is 5.50 x 104 cm2.

Find the electric flux through the surface.

2- Two point charges q1 = + 9.5 nC, and q2 = - 9.5 nC are separated by 5.5 mm, forming an electric dipole. If the torque exerted on the dipole has magnitude of 7.2 X 10 -9 N.m, (a) find the magnitude of electric dipole moment, (b) what is the electric field if the charges are in a uniform electric field whose direction makes an angle of 36.90, (C) what is the electric potential energy of this electric field ?

Explanation / Answer

Electric flux through the surface

Flux = EA cos x = 250* 5.5*10^4 * cos 45= 97722718.24 Nm^2/C

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a.

Dipole moment,

M= qa= 9.5*10^-9* 5.5*10^-3= 52.25*10^-12 Cm

b.

T= ME sin x

7.2*10^-9= 52.25*10^-12* E sin 36.9

E= 229.5 N/C

c.

Electric potential energy

U= -M*E cos x= 52.25*10^-12* 229.5* cos 36.9

= - 9.59*10^-9 J

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Comment in case any doubt..goodluck

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