Part A = 6.58 * 10^-13 N
Part B = negative x
Part C = attractive
Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.
What is the direction of the electric force?
Is this force attractive or repulsive?
Part A = 3.08*10^-14
If a drop is to be deflected a distance d = 0.320 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop?Click for More...
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...