Recent content by Kaoi

Okay, this was just a curious question that's been bobbing around in my head recently. The formula for gravitational force between two masses is F_{g} = G \frac{m_{1}m_{2}}{r^2} right? So, gravitational force increases as you approach the center of the earth. I was wondering, what is the...

Yes, you've converted your power the wrong way.

Oh, I marked the battery backwards. :smile: I was thinking of electron flow instead of the "positive-negative" flow. But the way I see it, for example, across the top branch: -Charge leaves battery (positive terminal) with 27 J of potential energy to expend. -Charge expends 12 J of energy...

I think I see what you're saying... So: Diff. between the original potential and the 32 \Omega resistor is (27 - 12) = 15 V. Diff. between the original potential and the 1 \Omega resistor is (27 - 1) = 26 V. So does that means the difference is (26 - 15) = 11 volts across the capacitor, or...

So, if the capacitor acts like an open circuit, then it's a parallel circuit with two resistors in series on each branch, like the diagram I've attached? If that's true, then: \Delta V = \Delta V_{13} = \Delta V_{24} R_{13} = 32 \Omega + 40 \Omega = 72 \Omega R_{24} = 1 \Omega + 27...

A Bridge Circuit Capacitor [Solved] EDIT: Problem has been solved. Homework Statement "The circuit has been connected as shown in the figure for a 'long' time. (See attached diagram) What is the magnitude of the electric potential across the capacitor? Answer in units of V."...

Homework Statement (Note: This is the fourth part of a single question, but it's the only part I'm having trouble with. Don't worry, it's not as simple as Q/V. :smile:) "What is the potential difference between a point midway between the [circular, parallel] plates [which are 1.43 x 10-4 m...

That clears it up a lot for me. Sorry for making you have to go to so much trouble. o:) So: \left(q_{neg}+\sqrt{\frac{F_{e,f}r^2}{k_{C}}}\right) ^2 - \frac{F_{e,f}r^2}{k_{C}} - \frac{F_{e,i}r^2}{k_{C}} = 0 q_{neg} + \sqrt{\frac{F_{e,f}r^2}{k_{C}}} = \sqrt{\frac{F_{e,f}r^2}{k_{C}} +...

But the problem lies in which F_{e} to use, since there are different F_{e}s on the sides of the equation, and they're not equal.

So: q_{neg}^2 + 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} - \frac{F_{e,i}r^2}{k_{C}} = 0 q_{neg}^2 + 2r\sqrt{\frac{F_{e,f}}{k_{C}}}q_{neg} = \frac{F_{e,i}r^2}{k_{C}} x=q_{neg}, I know that, which would make: q_{neg}^2 + 2q_{f}q_{neg} = (q_{neg} + q_{f})^2 - q_{f}^2 and that almost works, but...

Changing the signs, I get basically the same equation with a couple of signs reversed-- that still doesn't help me much... q_{neg}^2 + 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} - \frac{F_{e,i}r^2}{k_{C}} = 0 That's a strange square to complete...

But if the second charge is negative, wouldn't subtracting it make it addition?

\frac{F_{e,i}r^2}{k_{C}q_{neg}} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}} - q_{neg} q_{neg}+\frac{F_{e,i}r^2}{k_{C}q_{neg}} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}} q_{neg}^2 + \frac{F_{e,i}r^2}{k_{C}} = 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} q_{neg}^2 - 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} +...

If I solve that, I get: F_{e,i} = \frac{k_{C}q_{pos}{q_{neg}}}{r^2} q_{pos}q_{neg} = \frac{F_{e,i}r^2}{k_{C}} And I can't solve for q_{pos} and q_{neg} at the same time... I'm not sure I understand what you mean by "substitution" here... Wouldn't the negative charge cancel out part of the...

Electricity, Electric Force, and Charge Transfer Homework Statement "Two conducting spheres have identical radii. Initially, they have charges of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the magnitude of the negative charge. They attract each...