Recent content by stylez03

Sorry I left out the actual question: If the circuitry can detect a change in capacitance of 0.300 pF, how far must the key be depressed before the circuitry detects its depression?

ooh E = \frac {KQ} {R^{2}} I was thinking about this one but I thought you just meant in this chapter. Q = 7.30 * 10^-19 ?

E = sigma / e_o ?

The only equation from this chapter that I could find that would be relevant and give you E would be: V_{ab} = E*d E = \frac {V_{ab}} {d} ?

Homework Statement In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in...

Homework Statement Find the electric-field energy density at a point which is a distance of 10.0 cm from an isolated point charge of magnitude 7.30 nC. Homework Equations u = \frac {1} {2} * e_{o} * E^{2} The Attempt at a Solution u = \frac {1} {2} * 8.85*10^{-12} * 7.30^{2}...

I'm sorry I'm still not fully understanding that. I see from the equation above, you have the sum of the volume of the two raindrops, which gives the 3rd term, you said that the volume is conserved, equating that to 4th term. From what I can understand, R_{new}^3} does not seem to be...

Okay, that was my original intuition that the radius will grow, and you said the volume is also conserved. Now that we know the radius will grow, how do I go about solving for the new R from knowing that each individual raindrop has a radius of r?

So the radius is still the same?

Two identical raindrops, each with radius and charge specified in part (A), collide and merge into one larger raindrop. What is the radius of this larger drop, if its charge is uniformly distributed over its volume? Would it just be 2times the R? I don't think it would be, but I'm not sure how...

R = \frac {KQ^{2}} {(m_{p}*v_{i}^{2})} R = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(1.67*10^{-27})*(1350)^{2}} R = \frac {2.30*10^{-28}} {4.50*10^{-24}} R = 0.000051 F = \frac {K*Q_{1}*Q_{2}} {d^{2}} F = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}}...

After I find the closet approach, you said to use Coulomb's Law to find the force. Is the R in here equilv to d in Coulomb's Law where: F = \frac {K*Q_{1}*Q_{2}} {d^{2}} and F is what I'm after in t his problem?

m_{p}*v_{i} = \frac {KQ^{2}} {R} Divide by m_{p}*v_{i} \frac {KQ^{2}} {(m_{p}*v_{i})*R} Multiply by R R = \frac {KQ^{2}} {(m_{p}*v_{i})} Sorry my algebra is a little rusty

Ah let's show each step: KE_{i}1 + KE_{i}2 = PE_{f} 2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R} 2(m_{p}*v_{i}) = \frac {KQ^{2}} {R} We divide out KQ^2 to isolate R \frac {2(m_{p}*v_{i})} {KQ^{2}} = {R} ??

When we count both, would it just be KE_1 + KE_2? 2*(\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}}) = R