Hi,
An interesting problem!
When going through a curve of radius r with speed v_1, you have a centripetal acceleration of v_1^2/r. Hence, according to Newton's law, you need a centripetal force of m v_1^2/r, where m is the mass of the car (this will cancel out later).
Now, suppose that...
Hi JordanGo,
As Mindscrape noted, your ODE corresponds to a damped harmonic oscillator with a nonlinear damping term. To be specific, the damping term is the coefficient \beta(x^2-1). If this coefficient is positive, then the damping force will be in the opposite direction of the speed...
Hi Telemachus,
I believe you are right to doubt your procedure, since you are obtaining the potential by positing a surface charge, whereas actually the question is to determine the surface charge from the field.
My approach would be as follows. First, solve for the potential V(r) using...
Hi Lancelot59,
Although some of your intermediate expressions don't make sense (you can't have an expression like "\theta+a" because \theta is in radians and a is in meters), I do concur with your expression for V(r,\theta). (In the notation I'm used to, your constant k=1/(4 \pi \epsilon_0)...
Hi,
As Cipherflak says, we can find the answer by logical thinking. It is given that V(r) has a parabolic form, so we posit it has the form V(r)=ar^2+br+c where a, b, and c are constants to be determined.
Since we have three unknown constants, we need three conditions. These conditions...
Hi,
Indeed, vela, I had forgotten about that technique. Upon investigation I find that it does not make the problem much easier, however. By using the integrating factor technique, I recast the first-order ODE for y(s) into the following form...
Hi,
Specifically, your field component in the x-direction, E_x, is given by the formula E_x(x)=\frac{1}{\epsilon_0}\int_{d/2}^x{\rho(x')dx'} where \rho is the piecewise defined function given in the previous post. Choosing the lower limit at d/2 ensures the boundary condition mentioned...
Hi Bassalisk,
Thanks for reminding me to check the solution. My solution is indeed equivalent with the one you give, which can be verified by substituting z=d/2 into my equation (or x=z+d/2 into your equation).
Although my method using the integral form is the standard procedure taught in...
Hi Applejacks,
I verified your calculations, and I think you are on the right track. The -1 does indeed preclude the use of separation of variables to solve the ODE for y(s). If I put that ODE into Maple I get a complicated expression involving the error function, so it does not look like...
Hi Bassalisk,
Here is how I would solve the problem.
Firstly, I would change the coordinate system in order to exploit the symmetry of the problem. Defining the variable z=x-d/2, the charge density becomes:
\rho(z)=A\left[\left(\frac{d}{2}\right)^2-z^2\right].To solve the electrostatics...
Hi,
So you want to solve the second-order differential equation y''+3y'=\cos(t)-2. The corresponding homogeneous equation is y''+3y'=0, the solution of which is C_1+C_2\exp(-3t), where C_1 and C_2 are undetermined coefficients. The forcing function \cos(t)-2 can be expressed as the sum of two...