Anybody good in physics? Specifically Electrical Potential?

or lkobriensemail@gmail.com

Could you explain in a little more detail what you are having trouble figuring out?

I know it’s normally frowned upon to ask homework questions…..but I think study questions aren’t so bad…..

On capacitance: there should be some very straightforward equations that relate plate area, plate separation, and dielectric constant to capacitance. Once you have the capacitance there should be another straightforward equation that relates voltage to charge stored.

I know and I usually don’t answer homework questions simply because its homework. Unfortunately this class doesn’t even have homework and our grade is solely based on 2 quizzes, midterm, final.

For the first one I don’t understand how you found the potential energy (in either unit) with the only information given is V. Also how does a proton differ from an electron here?

The last 3, I’ve tried to just plug in numbers, but am still not getting the correct answers.

For first one, shouldn’t I have q..The only equation I know is U=Vq…....So I have U=(563V)(q)

Two unknowns?

A proton has a unit electrical charge of +1e or 1.60217646 × 10^-19 coulombs rather than being negatively charged of the same magnitude. It’s been quite a long time since I’ve done charged particles in an electric field so I can’t really say right off the top of my head exactly what to do. In this case I believe the q may be the charge of the proton in the electric field.

As for the other three, there doesn’t seem to be much more I can tell you. Just be careful of units (I’d put everything into base units first before doing any operations) and make sure you have your equations right.

Okay, lets analyse the question step by step.

The first part of question asks about the electric potential energy at a point.
The basics say,
“Electric potential at a point is the potential enegy energy spent to bring an unit positive charge from infinity to that particular point .”

Potential of an Electric field is Kq1/r and electric potential energy is Kq1q2/r where r is the distance between 2 charges.

If you place a proton at point P we get the it’s electric potential enegy is q2*electric potential
now q2 is the charge of the proton which is a unit positive charge and opposite to that of the charge on a single electronl which is 1.602*10^-19

Hence the Potentail enegy of the Proton will be 1.602*10^-19 * 563 = 563 electron Volt
I electron volt is = 1.602*10^-19 Joules so,
In terms of joules you could multiply the entities .602*10^-19 * 563 and get your answer.

Same is the answer for an electron as it carries the same charge as that of a proton.

2. The second part of the question is even more easier to resolve.
The electric potential =kq/r and electric field is kq/r^2 so electric potential is electric field*r

Electric filed is given as 1.05 106 N/C.

and the distance between the plates is given as as 1.05 mm =1.05 *10^-3 meter

So Electric filed potential is 1.05 106*1.05 * 10^-3

Now the charge is C has a formula eA/d where e is the permitivity of free space A is the area and d is the distance between the capacitor plates.A and d are given and the value of e 8.85418782 × 10–12 F/m.Now it’s just left to calculations.

3.

Total charge is equal to sum of charges on both the plates
On one plate the charge is +10.9 µC
Now the equation of capacitors say Q=C*V
C and V are given as 0.97uF and 51V

so, Q=51*0.97=49.47 uF so, the charge on the negetive plate is 49.47–10.9 = 38.57 uF

Now about the total capacitance,When capacitors are connected in series

1/C=1/C1+1/C2
=>C=C1C2/(C1+C2)

so total capacitance = 0.97*1.88/(1.88+0.97)—-> please calculate

4.The final part
In case of parallel plate capacitors with dielectrics

C=KeA/d where K is the dielectric strength given as 7, A is the area given as 0.1m^2
d given as 6mm and e is the permmitivity of free space as described above in second part of the solution.

Now capacitance could be found out by calculating by putting the vales in the equation stated above C=KeA/d.

Charged stored is easy to find out once you calculate the capacitance as charge Q=CV
Here C will be calculated in above quetion and V is given by the battery as 14 Volts so you get the charge.

Electric field is V/m which is the easiest of all to figure out.
here V is 14 and m is 0.6mm.So E=V/m

This solves each and every part of your question, but you have to do the calculations and find out the asnwers.For the sake of anonymousity,I can’t mail you ,but if you have any doubts in understanding any portion of the solution, please reply to this response and I’ll be happy to guide you further. Have a nice day.
ETW

@Shuttle128 Thanks, I realized this around midnight that night, I must have been having a major brain block :)

@engineeristerminatorisWOLV Just saw this response, so I pulled back out my book and your explanations helped me understand those problems so much more….Thank you!!