Recent content by Alan I

I'm working on the same problems and for the second one this is my solution attempt (Not sure that is right though): Because V1 = V2 = V you can work out Q1 in terms of r2 which is the variable you are looking for. → k[FONT=Georgia]Q1/r1 = k[FONT=Georgia]Q2/r2 (1) Then since [FONT=Georgia]Q...

Homework Statement A thin ring (radius r = 1.41 cm) carries a charge Q = 8.57 pC distributed uniformly along its length. The ring lies in the y-z plane, so the axis through its center is the x-axis . A small detector is moving along the positive x-axis toward the ring at velocity v = -0.543i...

Homework Statement Suppose a hollow metal shell (outer radius 25.3 cm, inner radius 5.2 cm) carries charge Q = -7.99 pC. There is a tiny hole in the sphere, so small that it does not affect the charge distribution or the electric field created by the charge. An proton is released from rest at...

Thanks, now I see it. So from Vdroplet = k (2.8x10-12)/r0, I got r0 = 4.27 x 10-5 Plugged that in V0 = 4/3*π*(r0)3 and got V0 = 3.27 x 10-13 Now Volfinal = 2V0 = 6.54 x 10-13 ⇒ 6.54 x 10-13 = 4/3*π*R3 ⇒ R = (1.56 x 10-13)1/3 ⇒ Vbig droplet = k [FONT=Georgia]Qtotal/R Thanks!

Homework Statement Suppose you have two identical droplets of water, each carrying charge 2.8 pC spread uniformly through their volume. The potential on the surface of each is 589 Volts. Now, you merge the two drops, forming one spherical droplet of water. If no charge is lost, find the...

potential difference between the two points: Vba = - ∫4.397.22 (1.2X-3.06) dx ⇒ Vba = -[1.2X2/2-3.06X]4.397.22 = -(9.184+1.870) = -11.054 ⇒-11.054 = Vb - Va Va=0 ⇒Vb = -11.054 V does that make more sense? :olduhh:

Homework Statement In a certain region of space, the electric field along the x-axis is given by: E = 1.2x - 3.06, where E is in N/C and x is in meters. If you set the electric potential equal to zero at (4.39 m,0), find the electric potential, in V, at the point (7.22 m,0). Homework...

How did you guys get this result because I did it over and over with the values provided and I got something else. Here's a summary of the values used according to my calculations: ro=0.0458 m r=9.1r0= 0.41678 m Q=5.02*10-8 C → from Vsurface dV/dr=-kq/r2 V/m ⇒dV/dr= (8.99*109) N*m2/C2 *...

OK thanks! I don't know how I messed up that algebra so bad. After re-checking my calculations this is what I got: r1 m = k N*m2/C2 * (3.6*10-9) C / 16.2 N*m/C ⇒r1 = 1.998 m r2 m = k N*m2/C2 * (3.6*10-9) C / 42.3 N*m/C ⇒r2=0.7651 m → smaller for the higher potential ⇒ r1-r2 = 1.23 m Now it...

OK I think I got it. So the mistake was the sign bc ΔU=-W=-∫F*dl , so since I got ΔU=-30J ⇒ W=30J.

Homework Statement A non-conducting sphere (radius 11.3 cm) has uniform charge density ρ = 0.596 μC/m3. Find the distance, in meters, between equipotential surfaces V1 = 16.2 Volts and V2 = 42.3 volts. (Distance is always positive.) Homework Equations V=kq/r ρ=Q/V The Attempt at a Solution...

Thanks!

Homework Statement A variable electric field permeates all space: E= (5.04x+ 84.2)x104i where x is in meters and E is in N/C How much work is done to move point charge q = 8.03 μC at constant velocity along the x-axis from point A at (3.63 m,0) to (7.67 m,0)? The sign will indicate who...

Homework Statement In a certain region of space, the electric potential along the x-axis is given by: V = 111x3 - 517x - 258 where V is in Volts and x is in meters. Find the force F (magnitude in N and direction) felt by a -444 μC point charge at point (56.5 cm,0). (For direction, all you...

I'm working on the same problem and this is what I got from following this thread: - the equality of final potential means: VAf=VBf ⇒ kqAf/rA=kqBf/rB ⇒qAfrB=qBfrA ⇒qAf=qBfrB/rA and qBf=qAfrB/rA (1) and then from conservation of charge: qAi+qBi=qAf+qBf ⇒qtotal=(qBfrA/rB) + qBf → solve for qBf...