Recent content by sept26bc

What is H?

I used E=vB to solve for be, but then I get: vEr2/kq and that leave E not solved now?

Homework Statement What are the units for electric energy density and magnetic energy density? Is it J/m^3 Homework Equations μ(mag)= (B2)/ (2μo) μ(electric) = (1/2) (εo) (E2) The Attempt at a Solution Magnetic energy density : T2/ N/A2 = J/m^3 electric energy density...

The entire question is stated as so: Two particles, each with electric charge q and mass m, are traveling in vacuum side-by-side, on parallel trajectories a distance d apart, both with speed v (much less than the speed of light). Calculate the ratio of the magnitude of the magnetic force to...

Thanks much everyone! the particles are moving on parallel trajectory, does this have any significance? And will the ratio be unit less?

Homework Statement There are two particles side by side in a vacuum with electric charge q and mass m, traveling at distance d apart, with v speed(less than speed of light). What is the ratio of magnitude of magnetic to electric force the particle exert on each other? Homework Equations...

the acceleration would be zero because its not moving.

well F=ma and E=F/q so E= ma/q |Felectric| is related to E by E=F/q and |Fgravity| related to g by F=ma correct? im not understanding why you have Felectric/ Fgravity=1

Vela. I'm really confused I've been working on this for hours. I know that the gravitational and electric force have to cancel in order for the particle to be hovering.how would another force be included in the equation q/m=a/E ? or am i using the wrong equation?

I don't understand the hovering part, would it be the gravitational force 9.81? what would I do then?

Okay Thanks! I found acceleration can you reassure my answer? so F=ma and E=F/q I substitute and get a= Eq/m and with the numbers plugged in I get: a= (-150N/C)(1.602*10^-19 Nm^2/C^2) / (1.67*10^-27 kg) a= -1.44*10^10 m/s^2 for charge to mass ratio I used E=ma/q rearrange to get q/m= a/E...

okay I fixed the units and decimal points on my paper. it would be -0.0654 C/kg . okay if that is correct how would I find the total energy associated with the electric field of the Earth, if Earth is a charged capacitor? Would I use U=1/2 CV^2. Does the value of capacitance of the Earth matter?

ma=Eq I divided by m on each side to get q/m= aE. Then a=-9.81 and E= -150. I multiplied. Oh wait okay so I was supposed to get: a/E= q/m. so It would be -9.81/-150 = -0.654. q/m = -0.654 Correct?

F= ma and E=F/q So i sub, and get ma= Eq then solved for q/m

I'm getting confused. So okay we have gravitational and electrical force both acting downward, so would the particle be negatively charged. and I am guessing this equation tells that "the magnitude of the force of gravity must equal the magnitude of the electric force": F=ma=Eq. But now what is...