I am confused about when a double integral will give you an area, and when it will give you a volume. Since we are integrating with respect to two variables, wouldn't that always give us an area? Don't we need a third variable in order to find the volume? Thanks for the help.
For Problem 2, just use the equation:
r= \frac{m*v}{q*B}
except that we just want the velocity perpendicular to the B-field, so we use the equation:
r= \frac{m*v*cos(\pi/6)}{q*B}
Not sure how to find the pitch though. I'm thinking maybe something like:
p = r*tan(pi/6)
If you...
I am trying to solve this problem:
The electric potential in a region of space is V = 210x^2 - 270y^2, where x and y are in meters. Find the E-field at (3m, 1m)
So I started with:
E = \frac{V}{d}
so then I plugged x and y into the electric potential equation and got
V= -200 Volts...
Alright, I see how it works now, thanks so much for the help. The final solution is:
V=\frac{Q*(\sqrt{R_{out}^2+z^2}-\sqrt{R_{in}^2+z^2})}{2*\pi*\epsilon_0*(R_{out}^2-R_{in}^2)}
Would it be something more like this?:
V=\frac{2Q}{(R_{out}-R_{in})^2}*\int_{R_{in}}^{R_{out}}\frac{dx}{2*\sqrt{x}}
I'm not sure if this is the right integral for the equation?
I am stuck on this problem:
I started this problem by looking at the electric potential of a ring, which is:
V=\frac{kQ}{\sqrt{R^2+z^2}}
So then if it varies in the thickness of the ring, would it be reasonable to have it be:
V=\frac{kQ}{\sqrt{(R_{out}-R_{in})^2+z^2}}
Or would I...
Well, since the tracks both start and end at the same heights, the speed must be the same if they both start at the same speed. Using the potential energy = mgh, and the mechanical energy formula K_f + U_f = K_i + U_i.
I am stuck on this Physics problem and desperately need some help.
As a science project, you've invented an "electron pump" that moves electrons from one object to another. To demonstrate your invention, you bolt a small metal plate to the ceiling, connect the pump between the metal plate and...