Recent content by secret5437

got it! thanks.

nope.

i'm trying to figure out how much charge runs through the capacitor. it has a capacitance of 27 microfarads. and the battery provides a potential difference of 118 V. I'm pretty sure the equation is C = Q/V which would be simplified to C x V = Q so that would be 27 x 118...wouldnt it?

Homework Statement The capacitor in Fig. 25-25 has a capacitance of 27 µF and is initially uncharged. The battery provides a potential difference of 118 V. After switch S is closed, how much charge will pass through it? (picture is in the attachment) Homework Equations C = Q/V U =...

Homework Statement (a) Find the equivalent resistance between points a and b in Figure P18.5 (R = 15.0 ). (refer to physics.gif for parts a. and b.) (b)Calculate the current in each resistor if a potential difference of 34.0 V is applied between points a and b. A (4.00 resistor) A...

and arrangement 2 looks like this: Q+......Q- Q+......Q- Arrangement 1 looks like this: Q+.....Q- Q-......Q+ (imagine those two arrangements as squares)

I know that the answer to Arrangement 1 is 0, because the two positive and two negative charges cancel each other out.

Homework Statement 3. Four charged particles are held fixed at the corners of a square of side s. All the charges have the same magnitude Q, but two are positive and two are negative. In Arrangement 1, shown , charges of the same sign are at opposite corners. Express...