Question

The potential in a region between *x* = 0 and *x*
= 6.00 m is *V* = *a* + *bx*, where *a*
= 19.4 V and *b* = -6.70 V/m.

(a) Determine the potential at *x* = 0.

V

Determine the potential at *x* = 3.00 m.

V

Determine the potential at *x* = 6.00 m.

V

(b) Determine the magnitude and direction of the electric field at
*x* = 0.

magnitude | V/m |

direction | ---Select--- +x -x |

Determine the magnitude and direction of the electric field at
*x* = 3.00 m.

magnitude | V/m |

direction | ---Select--- +x -x |

Determine the magnitude and direction of the electric field at
*x* = 6.00 m.

magnitude | V/m |

direction | ---Select--- +x -x |

Answer #1

In a certain region of space the electric potential is
given by V=+Ax2y−Bxy2, where A =
5.00 V/m3 and B = 8.00 V/m3.
Calculate the magnitude of the electric field at the
point in the region that has cordinates x =
2.20 m, y = 0.400 m, and z =
0
Calculate the direction angle of the electric field at
the point in the region that has cordinates x =
2.20 m, y = 0.400 m, and z =
0.

The electric potential in a region of space is V=( 260 x2− 160
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(in degrees) counterclockwise from the positive x-axis. THe
strenght of the electric field is 1200 V/m.

The electric potential in a region of space is V=( 260 x2− 160
y2)V, where x and y are in meters. What is the strength of the
electric field at (x,y)=(2.0m,2.0m)
?
What is the direction of the electric field
at (x,y)=(2.0m,2.0m)? Give the
direction as an angle (in degrees) counterclockwise from the
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The electric potential in a region of space as a function of
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An electron starts at rest at x = 0 and travels to
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Calculate the magnitude of the work done on the electron by the
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A particle travels with a velocity ?=6.00×10-5?―2.00×10-5? [m / s] and enters to a region where there is a magnetic field ?=0.500? ― 2.00? ― 1.40 ? [T]. Determine:
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Rank the following situations in order of their energy density,
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(i) a region of empty space where there is a...

A 6.70 −μC particle moves through a region of space where an
electric field of magnitude 1200 N/C points in the positive x
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If the net force acting on the particle is 6.21×10−3
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vx, vy, vz =
answer is 0,219,0 m/s
why is...

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