Recent content by marchcha

I'm a little confused on how to find the angle that the vector maxis to the x-axis.

Homework Statement A rectangular loop of wire with sides H = 22 cm and W = 77 cm is located in a region containing a constant magnetic field B = 0.82 T that is aligned with the positive y-axis as shown. The loop carries current I = 334 mA. The plane of the loop is inclined at an angle θ = 30°...

Ah so they'd be the same and you can say Q/C = I3R3 the equation above. Thanks!

Would it be just V?

Zero for both.

Ok so I3 is .0456A I don't understand how the capacitors charge has anything to do with the first loop though?

Ok, then how do you find I3?

So then Q = C (V - I1R1 - I4R4)?

So would I2 also be 0?

The capacitors current will be zero at steady state. I'm a little confused as to what this tells us about the loop current for that loop.

Homework Statement A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 41 Ω, R3 = 68 Ω and R4 = 154 Ω. The capacitance is C = 49 μF and the battery voltage is V = 12 V. The positive terminal of the battery...

Hey as a user of flipit physics I feel your pain. You treat it as two point charges because it is outside of the conductor.

what you are trying to do in this problem is act like the system is a point charge so take the lambda you found in the beginning and add it to the lambda given multiply it by two and k and then divide by the radius (0.081)

******length of one is not even needed disregard that

Ey(P) = 2*k(this is 9.0E9)*(Charge density of outer shell * area of outer cylinder + lambda of line) all divided by radius (.081m) let me know if this works out for you