Question

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 direction of the electric field at (x,y)=(2.0m,2.0m)? Give the direction as an angle (in degrees) counterclockwise from the positive x-axis. THe strenght of the electric field is 1200 V/m.

Answer #1

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
positive x-axis.

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 given by
V ( x,y,z ) = -x^2 + 2y^2 + 15. If a 5 Coulomb particle is placed
at position (x,y,z)=(-2,-2,0), what is the magnitude and direction
of the force it experiences?

In some region of space the electric potential is V(x) =
2sin(2x) + 2x. What is the electric field in this region? What
would an electron do if placed at x = pi/3 m, move left, right, or
stand still?

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...

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
direction, and a magnetic field of magnitude 1.25 T points in the
positive z direction.
If the net force acting on the particle is 6.21×10−3
N in the positive xx direction, find the components of the
particle's velocity. Assume the particle's velocity is in the x-y
plane.
vx, vy, vz =
answer is 0,219,0 m/s
why is...

A uniform electric field of magnitude 260 V/m is directed in the
positive x direction. A +12.0 µC charge moves from the
origin to the point (x, y) = (20.0 cm, 50.0
cm).
(a) What is the change in the potential energy of the charge
field system?
______ J
(b) Through what potential difference does the charge move?
______ V
Full, detailed solutions please!

The electric potential in a region of space as a function of
position x is given by the equation V(x) =
αx2 + βx - γ, where α =
2V/m2, β = 7V/m, and γ =
15V. All nonelectrical forces are negligible.
An electron starts at rest at x = 0 and travels to
x = 20 m.
Calculate the magnitude of the work done on the electron by the
electric field during this process.
Calculate the speed of the...

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
direction, and a magnetic field of magnitude 1.25 T points in the
positive z direction. If the net force acting on the particle is
6.24×10−3 N in the positive x direction, find the components of the
particle's velocity. Assume the particle's velocity is in the x-y
plane Find all three components and enter in Vx, Vy, Vz...

1. A proton is released in a region in space
where there is an electric potential. Describe the subsequent
motion of the proton.
2. In a certain region of space, the electric field is zero.
From this fact, what can you conclude about the electric potential
in this region? (a) It is zero. (b) It does not vary with position.
(c) It is positive. (d) It is negative. (e) None of those answers
is necessarily true.
please provide throughout explanation

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