An electric field of intensity 4.2kN/C is applied along the x-axis. Calculate th

Question

An electric field of intensity 4.2kN/C is applied along the x-axis. Calculate the electric fluxthrough a rectangular plane 0.350 m wide and 0.700 m long if thefollowing conditions are true. (a) The plane is parallel to the yz-plane.
1 Nm2/C
(b) The plane is parallel to the xy-plane.
2 Nm2/C
(c) The plane contains the y-axis, and its normal makes an angle of45.0° with the x-axis.
3 Nm2/C

(a) The plane is parallel to the yz-plane.
1 Nm2/C
(b) The plane is parallel to the xy-plane.
2 Nm2/C
(c) The plane contains the y-axis, and its normal makes an angle of45.0° with the x-axis.
3 Nm2/C

Explanation / Answer

   Electric flux through a plane is givenby            =   E* A * cos    E   =   Strength ofelectricfield,   A   =   Areaof plane   =   0.350 *0.700   =   0.245   m2,          =   anglebetween field and area vector  ( area vector is normal tothe plane of area)    a.      =   00             =   4.2* 103 * 0.245 * cos00   =   1.029 *103   N-m2/C    b.      =   900             =   4.2* 103 * 0.245 * cos900   =   0   N-m2/C    c.      =   450             =   4.2* 103 * 0.245 * cos450   =   7.28 *102   N-m2/C             =   4.2* 103 * 0.245 * cos00   =   1.029 *103   N-m2/C    b.      =   900             =   4.2* 103 * 0.245 * cos900   =   0   N-m2/C    c.      =   450             =   4.2* 103 * 0.245 * cos450   =   7.28 *102   N-m2/C             =   4.2* 103 * 0.245 * cos450   =   7.28 *102   N-m2/C

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