Part A = V = 1/(2πε0h) * ln(h/(2r) + sqrt(1 + (h^2 / (4r^2)))
Part B = V0 = q / (4πε0r)
What is the electric potential V at the origin?
What is the potential V0 in the limit as h goes to zero?
Part A = V(z) = 2kQ/a^2 * (sqrt(z^2 + a^2) – z)
Part B = E = 2kq/a^2 * (1 – (z)/sqrt(a^2 + z^2))
What is the electric potential V(z) on the z axis as a function of z, for z > 0?
What is the magnitude of the electric field E on the z axis, as a function of z, for z > 0?
Part A = F(V) = 1/2 * Aε0(V^2 / d^2)
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.Click for More...
Part A = B > A > (C = D) > F > E
In the figure there are two point charges, +q and -q. There are also six positions, labeled A through F, at various distances from the two point charges. You will be asked about the electric potential at the different points (A through F).
Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric potentials as larger than negative electric potentials.Click for More...
Part A = Option 3
A positive charge is brought close to a fixed neutral conductor that has a cavity. The cavity is neutral; that is, there is no net charge inside the cavity.
Which of the figures best represents the charge distribution on the inner and outer walls of the conductor?