Recent content by darwin59

I'm not sure how that would help. I could rearrange that as (-q/2a)(R2-a2)[1/(a+R)-1/(a-R)] to take care of the absolute value, but then I'll end up with (-q/2a)(R2-a2)/[(a-R-a-R)/(a2-R2), and I'm stuck with Q=-qR/a. Actually, come to think of it...that's what the charge of the image is...

We're assuming it's close enough to induce a total charge of -q on the sphere.

Well I worked it down further, and I got it to be +q, not -q, so I must be missing an extra negative somewhere. Here are my steps: ∫d\Phi from 0 to 2∏ is just 2∏ I used u substitution, with u=R2+a2-2aRcosθ, and du=2aRsinθdθ. So now I have Q=(-qR/2)(1/2aR)(a2-R2)∫u-3/2du...

Homework Statement A point charge +q is situated a distance a from the center of a grounded conducting sphere of radius R. Find the induced surface charge on the sphere as a function of θ. Integrate this to get the total induced charge.Homework Equations σ=-εo*∂V/∂r Q=∫σ daThe Attempt at a...

I figured out my problem. I was actually doing it right, but I'm using a different calculator, so I was inputing numbers wrong.

Homework Statement Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) = (-23.5 cm, 0) carries current I1 = 3.3 A in the negative z-direction. The wire at (x,y) = (23.5 cm, 0) carries current I2 = 1.2 A in the positive z-direction...

Nevermind, I got it! I don't know if it was the correct method, but I used the equation sinθ=D/R.

I'm having serious problems with this one. I can't seem to get equations that will help me out. All I've gotten is that the vector equation is <2.61e5*cosθ, 2.61e5*sinθ>. There's the circle equation, x^{2} + y^{2} = R^{2}. There's the force equation, F=\frac{mv^{2}}{R}. There's also s=Rθ...

I know that the proton will move in a circle since the velocity is perpendicular to the field, which I figured to be in the negative z direction. I used the components of the velocity to determine that the angle the proton leaves the field is about 32.55 degrees, which is also the angle from...

A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x.,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.7 m in the x-direction. The proton leaves the field...

Thanks for the hospitality! I'm not sure how I didn't try that out, or how exactly I got stuck down the path that the conducting slab should have a net charge, but I got it. Thank you very much SammyS!

Homework Statement An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.34 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.14 μC/m2 is located at x = c = 26 cm.. An uncharged infinite conducting slab is placed...