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)
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...