Recent content by David Day

Yeah, I was thinking that using the same amount of wire, the volume would be constant, which isn't actually the case. So it seems to me that if the wire is of length x, and the circumference of the wire is 2πrN for each uniform winding, then x = 2πrN and r = x/2πN. I'm not sure if this is...

Homework Statement [/B] 1. Two solenoids, A and B, are wound using equal lengths of the same kind of wire. The length of the axis of each solenoid is large compared with its diameter. The axial length of A is twice as large as that of B, and A has twice as many turns as B. What is the...

Homework Statement [/B] After how many time constants is the stored energy of a discharged capacitor one-fourth its initial value? Homework Equations [/B] (1) U = Q2/2C (2) U = Qmaxε/2 (3) q(t) = Qie-t/RC The Attempt at a Solution The solution can be correctly attained using the first...

Homework Statement Using Kirchhoff’s rules, (a) find the current in each resistor shown in Figure P28.31 and (b) find the potential difference between points c and f.[/B]Homework Equations [/B] Σ ΔV = 0 (KVL) The Attempt at a Solution [/B] I have been trying to set up a system of equations...

That really flicked a switch in my head. Thanks for that!

A problem I solved asks this: There is a point charge of 3q at the origin and another charge of -2q at x = 5. At what value of x relative to the origin is the electric field equal to zero? So: E = E1 + E2 0 = E1 + E2 E1 = -E2 kQ1/x2 = -kQ2/(5-x)2 Solving for x, the point at which these...

I have some understanding, but I'm not sure about how accurate it is: Electrostatic force is given by F = qE, where F and E are both vector quantities. If the dot product of either side and the displacement vector Δs along an equipotential line is taken, the equation becomes F⋅Δs = qE⋅Δs. F...

Ah, yes. I'm always forgetting to evaluate final - initial. Thanks for that.

As Vagn and jtbell say, I see now, using equation 2, that by calculating the change in potential energy of the ball initially at ground level and at a final position of 1 m above the ground, the result is the same as that given by equation 1, that is: Uf = -Gm1ME/(RE + 1) Ui = -Gm1ME/(RE) Uf -...

I forgot that g = GM/r2, so I understand a little better how the equations are related. As for my calculations: Ball: m = 1 kg h = 1 m With equation 1: U = mgh = (1 kg)(9.8 m/s2)(1 m) = 9.8 J With equation 2: -Gm1m2/r = - (G)(1 kg)(ME kg)/(RE + 1 m) = 6.26E7 J, where ME = mass of Earth =...

I am hoping to get a deeper understanding of the difference between two different gravitational potential energy equations, the first of which is given by U = mgh and the second given by U = (Gm1m2)/r I first assumed that in a system consisting of the Earth and, say, a tennis ball, these...

The question is stated as the following: When a 3.60 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.83 cm. What is the force constant of the spring? The correct answer was acquired by using the equation F = mg = -kx, where k is the spring constant and x...