Question

The electric potential of a charge distribution is given by the equation VKY.Z) = 2xy-372 -42where x,y,z are measured in meters and Vis measured in volts. Calculate the magnitude of the electric field vector at the position xyz)-(1.0, 1.0, 1.0). Select one: O a. 14.2 V/m b. 18.2 V/m O C. 12.2 V/m d. 10.2 Vim ОО e. 16.2 Vim

Answer 1

Solution The electric field vector is given by ² = -ov Now, since, v= - 2x3y² 3 4²23 + 4z2x² Therefore, we have È = -√(- 2x² 4 2 – 342 z2 +42²x²) => È =-(3++)(-2x3y2-3y?z? +42+x2) = = -21-2x3y2 3y2z2 + 4z2x2) - 2 2 - 2x3 y 2 – 3 4 2 2 3 + 4z2x²) ? -21-2x3y2 3y z2 + 422x2) > ] = (242278-0-4722x2) î +( 23°372 +32° 3,42-0) +(0 +34? :23-43?322) Әх

= E = ( 2 y2 3x²_ 422 2x) ? +(2x3 2y + 37 +32²29) +(3y2 322 - 43222) > È = (642x²8 x 2² i + (4 x3y + 69 23 +(9y2z² – 8x² z) Therefore, the electric field vector at the position (X,Y,Z) = (1.0, 1.0,1.0) is È = [6 (1.0)2 (1.0)2 – 8(1.0) (1.0)2] ! + [4 (1:0)3(1-0) + 6(1.0) (1-0)3) + 9 (1.0) 2 (1.0)2 – 8 (1.032 (1.0)] ĥ → = [6-8]{+(4+6] 9 + (9-8] R = -2 + lo +10 ŷ ti = È = -2² +10 ht - tik 介 企