Day 16 Potential of Continuous Charge Distributionsb

At the beginning of the class, we first started off by estimating the potential from a charged ring. We got that V = kq/r and r is square root and a^2 + b^2 in this case.

In this case, r is sqrt(x^2+4a^2), another is sqrt(2a^2+x^2)

We also found the angle's relationships with a and x. 

We plugged in the previous equation and got V(θ) = kqcosθ/x

We also used excel to find the total potential energy of the charged ring, which the ring is divided into 16 parts. To find the total potential energy, we need to added everything up.

In this picture, we did the actual calculation for calculating the total potential energy. And the result is the same as the excel's.

We used the equations in the picture to find the equation of Ex = KQx/(x^2+a^2)^3/2

We also found that ΔV = KQ/sqrt(x^2+a^2)

We did a problem to calculate the electric potential due to a finite length line-charge.

We used the equation we derived previously and calculated the electric potential.

Then we used excel to calculate the same result.

These three pictures shows the equipotential surfaces where we drew perpendicular lines.

Electric Potential Lab




The pictures above are the equipment we used in this experiment. We used them to measure different voltages at different positions to calculate the Δv/Δx.

Pictures above are the data we recorded during the experiment.

Finally, we used excel to find Δv/Δx.

Conclusion

In today's class, we did various problems on finding electric potential in different cases. We also learned that using excel would be an easier and faster way if there are too many parts in a charged bar. We also learned about equipotential surfaces and did a lab on electric potential.

Day 15 Power and Potential

At the beginning of the class, the professor wanted us to find the arrangement of 2 light bulbs and 2 batteries to make the light bulb to bright the brightest and dimmest. The brightest is the top arrangement and the dimmest is the bottom arrangement.

In these two pictures, the professor asked us to list three ways to find ΔT. So, by the equation 

Q = cmΔT and Q = Pt, in order to find ΔT, we need to find time, mass of water, and the power.

In these two pictures, the professor used the logger pro to graph temperature vs. time as we gave the voltage to the water. The blue graph has 2 times of the voltage of the red one. However, the slope is not doubling but increase by a factor of 4 since P = IV, V = IR -> 2V = 2IR -> P = 4IV.

We also did a problem to find the current by the equation I =V/R.

Here is another work problem that we found that work in electric field is path independent, which means electric field is conservative.

These pictures shows an activity we did which the work done on a charge traveling in a uniform electric field. Different path has different work due to different angles.

This picture shows a particle and its electric field, which looks like circles.

We derived the electric potential energy to be kQ/r

These two pictures shows the code for VPython assignment for calculating the electric potential and electric field using VPython.

Lastly, professor told us to identify the difference between plagiarism and collaboration.

Conclusion

In today's class, we first learned how to make bulbs shine the brightest and learned the circuit diagrams. We also did some activities on work in electric field, electric potential energy and finally we discussed the difference between plagiarism and collaboration.

Day 14 Current Voltage and Resistance

At the beginning of the class, we first watched a video about lighting a light bulb

Then, we did an activity called arrangement that cause light and made 3 arrangements with light bulb, battery, and a wire that cause light, and 2 that do not cause light.

We also used two batteries to make the bulb lit twice brighter.

In this picture, the professor is using a rubber bar to rub by animal skin and put the bar on the ball, we discovered that two pieces of sheet-metal were separate.

In this picture, we used the battery to touch the ball, but nothing happened.

 We made a simple electric circuit and defined that battery give energy, energy is getting used up, and we need a wire because we want the energy to go back from the bulb to battery. And the reason why the bulb would light is because there is a circuit between the battery and the bulb that have the equilibrium of charge.

We also discussed the concept of electric potential difference of voltage. 

In this picture, we used the equation Power = V*I to calculate power to be 0.18 J/S.

In the pictures, it is the model C for current flow which there will be less current in the return wire than the coming wire.

We did an activity called measuring current and we measured that the value of the current in amps to be 132 mA in positive direction and 130 mA in the opposite direction

In the picture, we did a problem and calculated vd (drift velocity and current) by the equation

 I = ρnqvdA. 

Next, Ohm's law was introduced to us and we drew a I vs. V graph for V = RI where R is ohm. And this equation is only true when we have a good wire.

In these pictures, the professor used logger pro to graph I vs. V graph.

This is the I vs. V graph. It is a linear line.

In this picture, we found that unit of R which is J*S/C^2 which is also called ohm (Ω).

Then, we did an example problem about resistivity. And we found that the resistivity is proportional to the length and the less resistance the more electric conductive.

Finally, in the picture, we found that R is inversely proportional to cross section area of the wire, and we found that R = P*L/A 

Conclusion

In today's class, we first discussed about how to light a light bulb in various ways with batteries and wires. Then we talked about the relationship between voltage, current, and power. And finally we were introduced the ohm's law and have some applications and problems on ohm's law.