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In a certain region of space the electric potential is given byV=Ax2y-Bxy2, whereA=5.00V/m3andB=8.00V/m3. Calculate the magnitude and direction of the electric field at the point in the region that has coordinates x = 2.00 m, y = 0.400 m, and z = 0.

Short Answer

Expert verified

E=-6.72i-7.2jN/CE=9.85N/C=227

Step by step solution

01

parameters.

V=Ax2y-Bxy2A=5.00V/m3B=8.00V/m3x=2.00my=0.4m

02

Calculating the magnitude and direction of the electric field.

The electric field is given by

E=-v=Ex$+Ey$

The electric field is equal to the negative gradient of the potential so

Ex=-Vx=-2Axy+By2

Also in y direction

Ey=-Vy=-Ax2+2Bxy

The electric field is given by

E=[-2Axy+By2]i^+[-Ax2+2Bxy]j^

The electric field E(2.0,0.4,0) is given by

localid="1664280297072" E2.0,0.4,0=-2520.4+80.42i$+-522+280.42=-6.72i$-7.2j$NIC

The magnitude E I given by

E=(7.2)2+(6.72)2)=9.85N/C

Since both components are negative the angle lies in the 3rd quadrant so add to magnitude of angle

=180+arctan13.6/6.72=227

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