_______ V b.) An object (point mass m=3 kg, q=+6 C) is now placed at the origin.

Question

_______ V

b.) An object (point mass m=3 kg, q=+6 C) is now placed at the origin.

What is the voltage at the origin?
V = ______ V

How much energy does the object store here?
U = _______  J

c.) The object is given an ever-so-slight nudge in the -y direction.

Use your electric field function to show that the object will always feel a force that points along the -y axis.

When the object arrives at y=-2 m, find:

the voltage at this location: _______ V
the energy stored by the object: _____ J
its speed: ______ m/s

Suppose the following function describes the electric potential on the xy plane: V(x,y)=x^2 - y^2 + 1 (Units omitted; you may assume [x], [y] = m and [V] = V.) a.) Find a general function j Sketch the xy plane with the local fields you found indicated at their respective locations. Then answer: What is the common electric potential at all of the coordinates listed above? _______ V b.) An object (point mass m=3 kg, q=+6 C) is now placed at the origin. What is the voltage at the origin? V = ______ V How much energy does the object store here? U = _______ J c.) The object is given an ever-so-slight nudge in the -y direction. Use your electric field function to show that the object will always feel a force that points along the -y axis. When the object arrives at y=-2 m, find: the voltage at this location: _______ V the energy stored by the object: _____ J its speed: ______ m/s i hat + _______ N/C vector E(-4 m,-3 m) = ______ N/C j Point D: i hat + ______ N/C vector E(-4 m,+3 m) = ______ N/C j Point C: i hat + _______ N/C vector E(+4 m,-3 m) = _______ N/C j Point B: i hat + _____ N/C vector E(+4 m,+3 m) = ____ N/C vector E(x,y) and then use it to find the electric field at points A-D, which are at the indicated (x,y) coordinates: Point A:

Explanation / Answer

E = -dV/dr = -(2x i - 2y j)

a. E(4,3) = -(8i-6j)

E(4,-3) = -(8i+6j)

E(-4,3) = (8i+6j)

E(-4,-3) = (8i-6j)

Common Electric Potenital = (42) - (32) + 1 = 8 V

b.

V(0,0) = 1V

U = Vq = 6J

c.

F = qE

E(0,0) = 0

E(0,-y) = -2y j (in negative y direction)

F = -12y j

Now, Force = ma = - m d2y/dt2 and equal to -12y.

Solve for y and get velocty (byt diff y wrt t). Use intital condition: t=0 => y=0.

V(0,-2) = -3V

U = KE + PE = (1/2)*m*v2 + qV = (1/2)*m*v2 - 18 J

Cheers

PS - Ask if you have any doubt.

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