Recent content by 743344

Homework Statement There is a current of 0.25 A in the circuit of the figure What is the value of the resistance R? Homework Equations V = IR The Attempt at a Solution since the circuit is in series: V = 6+ 12 = 18V Req = R + 6 + 12 V = IReq Req = 18/0.25 = 72...

ya i have no idea how mastering physics got Q1...but i eventually got Q2/3... i used Ceq (taking all the circuits in series)... weird question

then how would you find the charge across each capacitor...?

sorry i forgot to mention the question ask to find Q1, Q2 , Q3 and V1, V2 , V3 what i am thinking this is my understanding so far 1) while the switch is in position A...C1 would be charged to Q = CV = 15(100) = 1500 microC 2) once the switch is in position B...the charge on C1 is...

Homework Statement Initially, the switch in the figure View Figure is in position A and capacitors C_2 and C_3 are uncharged. Then the switch is flipped to position B. Afterward, what are the charge on and the potential difference across each capacitor? Homework Equations...

what is the charge on the point charge F = qE W = Fdcos(angle) W = qEd U(electric potential energy ) = - W(electric) then U = -qEd

Homework Statement Two 10-cm-diameter metal plates are 1.0 cm apart. They are charged to \pm12.5 nC. They are suddenly connected together by a 0.224-mm-diameter copper wire stretched taut from the center of one plate to the center of the other. a) What is the maximum current in the wire...

thank you

[SOLVED] electric field given electric potential Homework Statement The electric potential in a region of space is V = ( 210 x^{2} - 170 y^{2} ) V, where x and y are in meters. What is the strength of the electric field at ( 1.00 m, 3.00 m ) ? Homework Equations E = -dV/ds...

Homework Statement A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length. What is the electric potential at the center of the cylinder? Homework Equations n = Q/A V = kQ/r The Attempt at a Solution 1) the cylinder has uniform...

ok so if dF= Ls*dx*E dF=Ls*dx*(k2lamda)/x sum (dF)= sum (Ls*dx*(k2lamda)/x) F = Ls*x*(k2lamda)/x but x cancels how can i find the integration?

ok let's try this 1. charge/unit length = lambda = 18nc/0.06 isn't this the linear charge density of the plastic stirrer? 2. F = dqE where E = (k2lamda)/dx then F = dq(k2lamda)/dx 3. integration F = dqk2lamda(1/dx) |from 0.02 to 0.06am i right so far? for lamda which linear charge density...

ok so this is what i have E(long wire) = (k2|lamda|)/r Fnet = q(point charge)E then Fnet = q((k2|lamda|)/r where r is the distance from the wire = x = Fnet = q((k2|lamda|)/x (in the direction away from the wire (which is right in the diagram, a repulsion force)) but then how do i find x...

this is what i tried: (Ex) = kQ/r^2 where r = the distance from the the wire to a distance x on the rod r = 0.02 + 0.06 - x if d = 0.02 and L = 0.06 then E = kdeltaQ/(d+L-x)^2 deltaQ = lamda/delta x then E = k(lamda/delta x)/(d+L-x)^2 E = sum of Ex = k(lamda/delta x)/(d+L-x)^2 intergrate E...

Homework Statement Your physics assignment is to figure out a way to use electricity to launch a small 6.0-cm-long plastic drink stirrer. You decide that you'll charge the little plastic rod by rubbing it with fur, then hold it near a long, charged wire. When you let go, the electric force...