What is the radius of the equipotential contour corresponding to V = 1180 V and

Question

What is the radius of the equipotential contour corresponding to V = 1180 V and y = 5.00 m?

The answer is 43.3m, but I don't know how to get the answer. Steps online say to do (V/A +3y2)1/2, but that doesn't get me the answer.

Reference:

The electric potential V in a region of space is given by the following expression, where A is a constant.
V(x, y, z) = A (2x2 - 6y4 + 2z2)

(a) Derive an expression for the electric field vector E at any point in this region. (Use the following as necessary: A, x, y, and z.)
x-component
Ex = -4Ax
Correct: Your answer is correct.
y-component
Ey = 24Ay3
Correct: Your answer is correct.
z-component
Ez = - 4Az

(b) The work done by the field when a 1.90 µC test charge moves from the point (x, y, z) = (0, 0, 0.320 m) to the origin is measured to be 8.00 10-5 J. Determine A.
A = 206 V/m2

(c) Determine the electric field at the point (0, 0, 0.320).
E with arrow(0, 0, 0.320) = 263 N/C

Explanation / Answer

(a)

Given:
V = A (2x2 - 6y4 + 2z2)

Electric field is the negative gradient of the electric potential function. The negative exists, because a positive test charge will be forced to travel to a lower electric potential, by our reasoning for defining electric potential.

E = -V

The upside-down delta represents a vector of partial derivatives:
= <d/dx, d/dy, d/dz>

Thus:
E = <-dV/dx, -dV/dy, -dV/dz>

Carry out partial derivatives, remembering that other spatial coordinates are constants if not in the Leibniz notation denominator.
Ex = -dV/dx = -4Ax
Ey = -dV/dy = 24Ay3
Ez = -dV/dz = - 4Az

The vector E =-4Ax i +24Ay3 j - 4Az k

(b)

The potential energy of q = 1.90x10^-6 C , is U = q V ;
the work done by the field when q moves from P to O is :
W = - [U(O) - U(P)] = U(P) - U(O) = q[V(P) - V(O)] = qA*0.3202.
Then A = 8x10-5 / (1.90x10-6 x0.3202) = 411.18 V/m^2

(c)

The vector E =-4Ax i +24Ay3 j - 4Az k

x = 0 , y = 0 , z = 0.320m

E = sqrt(02 + 02 + (4*411.18*0.320)2))

E = 526.310

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.