1) In a certain region, the electric potential due to a charge distribution is g

Question

1) In a certain region, the electric potential due to a charge distribution is given by the equation V(r.yz)-3ry2 + yz'2'x, where x, y, and z are measured in meters and V is in volts. Calculate the magnitude of the electric field vector at the position (r.y.z)-(1.0,1.0, 1.0). A)-8.1 V/m B)74 V/m C)4.3 V/m D) 8.6 V/m E) 2.0 V/m 2) The graph in the figure shows the variation of the electric potential V (measured in volts) as a function of the radial direction r (measured in meters). For which range or value of r is the magnitud of the electric field the largest? A) from r-0 m to r-3 m B) from r = 3 m to _4m C) atr-4 m D) at r-3 m E) from r- 4 m tor-6m 3) In a certain region, the electric potential due to a charge distribution is given by the equation V(x,y)-2xy-x2-y, where x and y are measured in meters and Vis in volts. At which point is the electric field equal to zero? B) x = 0.5 m, y = 0.5 m C)x-0.5 m, y-1m D)x-1 m, y- 0.5 m

Explanation / Answer

1) given
V = 3x^2y^2 + yz^3 - 2z^3x
hence
E = -dV/dx i - dV/dyj - dV/dz k
hence
E = -(6xy^2 -2z^3) i - (6x^2y + z^3)j -(3yz^2 - 6z^2x) k
at (x,y,z) = (1,1,1)
E = -4i -7j + 3k
hence
|E| = 8.602 V
hence option D)

2) electric field is proportional to the slope of V-r graph
slope os greatest for
3 to 4 cm
hence
option B)

3) V = 2xy - x^2 - y
for E = 0
dV/dx = 0
dV/dy = 0
dV/dz = 0

hence
2y - 2x = 0
=> x = y

2x - 1 = 0
=> x = 1/2
hence
y = 1/2

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