Question

a) An electric field has the expression E = 3.0î + 2.0ĵ V/m. The position of...

a) An electric field has the expression E = 3.0î + 2.0ĵ V/m. The position of a particle is given by ds = 5.0dyĵ m. Determine the electric potential of this particle from x = 2.0 cm to x = 10.0 cm.

b) The voltage across an air-filled capacitor is measured to be 25.0 V. An unknown dielectric is inserted between the plates, the voltmeter now reads 10.0 V. What is the most likely material this dielectric is made out of?

c) The plates of this capacitor are separated by 1.0 mm. The maximum charge that can be stored in this capacitor is 30.0 μC. Determine the area of the plates of this capacitor.

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 ×× 1077 V/mV/m. A piece of polystyrene...
Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 ×× 1077 V/mV/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. A) When the electric field between the plates is 79 %% of the dielectric strength, what is the energy density of the stored energy? B) When the capacitor is connected to a battery with voltage 500.0 VV, the electric field between the plates is 79 %% of the dielectric...
Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 × 107 V/m. A piece of polystyrene...
Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 × 107 V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. Part A When the electric field between the plates is 76 % of the dielectric strength, what is the energy density of the stored energy? Express your answer with the appropriate units. u = nothingnothing SubmitRequest Answer Part B When the capacitor is connected to a battery with voltage...
Consider a parallel plate capacitor with plate dimensions of 10 cm x 10 cm and a...
Consider a parallel plate capacitor with plate dimensions of 10 cm x 10 cm and a gap spacing of 2.00 cm. The capacitor is connected to a battery supplying a voltage across the capacitor of 2.00 kV. After the capacitor is fully charged, the battery is disconnected. A dielectric material is then inserted into the vacuum space to completely fill the capacitor. This result in a decrease of the capacitor voltage to 1.00 kV. a.Find the vacuum electric field between...
An initially uncharged air-filled capacitor is connected to a 4.65 V charging source. As a result,...
An initially uncharged air-filled capacitor is connected to a 4.65 V charging source. As a result, 3.35 × 10 − 5 C of charge is transferred from one of the capacitor's plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant κ of this substance is 7.05 . Find the voltage V across the capacitor and the charge Q...
An initially uncharged air-filled capacitor is connected to a 2.01 V charging source. As a result,...
An initially uncharged air-filled capacitor is connected to a 2.01 V charging source. As a result, the capacitor acquires 5.49×10−5 C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant ? of this substance is 7.11. Find the voltage ? across the capacitor and the charge ?f stored by it after the dielectric is inserted and the circuit has returned...
An initially uncharged air-filled capacitor is connected to a 5.15 V charging source. As a result,...
An initially uncharged air-filled capacitor is connected to a 5.15 V charging source. As a result, the capacitor acquires 7.19×10^−5C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant ? of this substance is 7.39. Find the voltage ? across the capacitor and the charge ?f stored by it after the dielectric is inserted and the circuit has returned to...
An initially uncharged air-filled capacitor is connected to a 4.59 V charging source. As a result,...
An initially uncharged air-filled capacitor is connected to a 4.59 V charging source. As a result, the capacitor acquires 9.05×10−5 C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant ? of this substance is 7.37. Find the voltage ? across the capacitor and the charge ?f stored by it after the dielectric is inserted and the circuit has returned...
1) Find the electric field inside a 10 microFarad parallel plate capacitor when connected to a...
1) Find the electric field inside a 10 microFarad parallel plate capacitor when connected to a 6V battery if the gap between the capacitor plates is filled with air. Repeat your calculation if the gap is filled with paper with a dielectric constant of 4? 2) A capacitor is formed between two metal plates, both 10 x 10cm, separated by 1mm. A) What is its capacitance? B) When connected to a 10 V battery how much charge flows onto the...
The electric field between the plates of a paper-separated (k=3.75) capacitor is 8.245104 V/m. The plates...
The electric field between the plates of a paper-separated (k=3.75) capacitor is 8.245104 V/m. The plates are 1.95 mm apart, and the charge on each plate is 0.775 C. Determine the capacitance of this capacitor and the area of each plate.
If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its...
If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is at the maximum magnitude that the electric field can have without breakdown occurring. The dielectric strength of air is 3.0x106 V/m and that of neoprene rubber is 1.2x107 V/m. A certain air gap parallel plate capacitor can store no more than 0.075 J of electrical energy before breaking...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT