i would try using the plane i outlined with red arrows for your area.
Flux= E x dA cos(theta), where cosine is the vector that defines the area of the plane. usually, this vector is normal (perpendicular) to the surface of the plane. when the area vector and the electric field are parallel...
this derivative is really fouling me up here. this time, i used the product rule, and i got
-9/2zKQ(z^2+a^2)^-5/2+KQ(z^2+a^2)^-3/2
then i factored out a KQ(z^2+a^2)^-3/2 to get
KQ(z^2+a^2)^-3/2(-9/2z((z^2+a^2)^-1)+1)
after that, i broke it apart into factors and set each factor equal...
your attachment isn't working, but i would assume it is bent in the middle forming a triangle with 2 parts being .05m. find the hypotonuse of that triangle, and use that and the part of the square remaining straight to form a new area. the flux will be parallel to the vector that defines the...
this is the second part of a 3 part question the first part was:
A uniform circular ring of charge Q=4.40 microCoulombs and radius R=1.30cm is located in the x-y plane, centered on the origin as shown in the figure. What is the magnitude of the electric field E at point P, located at z=3.30...
i guess that was right. i got the problem right, so that should help me on a few other problems. i will probably be back here asking more questions later though!
thanks for the welcome!
i have been thinking about it on the next homework problem actually, and i think it has something to do with conservation of momentum. something like:
1/2(mv^2)+mgy= 1/2(mv^2) +No potential at the bottom.
the initial velocity is 0, so the first initial kinetic...
Homework Statement
Tarzan, who weighs 668 N, swings from a cliff at the end of a convenient vine that is 23.0 m long (see the figure). From the top of the cliff to the bottom of the swing, he descends by 3.2 m.
If the vine doesn't break, what is the maximum magnitude of the tension in the...