Recent content by islimst3ri

Since the electric field is moving with the vector in the direction of j-hat, it is moving upwards, while the gravitational field, intuitively, is moving downward. Since the electron was able to travel vertically for 1.2 µm, this indicates that the E-field is stronger than the g-field, thus this...

Tim, So WebAssign is down right now thus I can't check my answer. But I'll show my work and could you please scan over it to see if its right. A simple yes or no will suffice. WE-Field = KE = qEh = -(1.6E-19 C)(-1.8E-10 N/C)(1.2E-6 m) = 3.46E-35 J Wgravity = mgh = (9.1E-31 kg)(9.81...

When I follow this approach, I have get W = KE = qEh = (1.6E-19 C)(-1.8E-10 N/C)(1.2E-6 m) = -3.46E-35 J After finding work, I plugged it into W = 1/2 mv2 as follows: -3.46E-35 J = (.5)(9.1E-31 kg)(v2) -6.91E-35 J = (9.1E-31 kg)(v2) -7.60E-5 J/kg = (v2)...

Dick, Maybe I'm missing something here but I don't see it, what do you mean by using the conservation of energy? Are you suggesting that I should use the kinetic form, by setting mgh = 1/2 mv^2? If so I don't understand why the Electric field charge could be involved at all. Please explain a...

This is the question: "An electron is released from rest in a weak electric field given by vector E = -1.80 10-10 N/C jhat. After the electron has traveled a vertical distance of 1.2 µm, what is its speed? (Do not neglect the gravitational force on the electron.)" So far, the only approach...