Question

Please write your work neatly so I could understand it

3. A very long, straight line of charge runs parallel to a very large, planar sheet of charge. The 2 charge density of the line is -2.0 nC/cm and that of the sheet is +0.50 nC/cm^2 . The distance

between the line and the sheet is 8.0 cm. An electron is released at rest, at a location that is halfway between the line and the sheet. Use conservation of energy to calculate the speed of the

electron when it strikes the sheet. (Hint: calculate the potential difference due to line only and then the potential difference due to the sheet only, and add them up to get the potential difference

you need to determine the speed)

4. You have three capacitors, of 10 nF, 20 nF and 30 nF capacitances, and a voltage source of 11

Volts.

A. Connect the capacitors in series and apply the voltage across the network. Calculate the equivalent capacitance of the network, the charge stored in the network, the energy stored in the

network, and the voltage across each capacitor.

B. Connect the capacitors in parallel and apply the voltage across the network. Calculate the equivalent capacitance of the network, the charge stored in the network, the energy stored in the

network and the charge stored in each capacitor.

Answer 1

3.

Linear charge density of line of charge, λ = - 2.0 nC/cm = - 2.0 x 10-7 C/m

Surface charge density of sheet of charge, σ = 0.50 nC/cm2 = 0.50 x 10-5 C/m2

Distance between line of charge and sheet of charge, R = 8.0 cm = 0.08 m

Distance from mid point to each charge distribution, R/2 = 0.04 m

Electric field at a point near a sheet of charge, E = σ/2εo

We have, E = - dV/dr

Potential difference , dV1 = - Edr = σ/(2εo ) x R/2 = 0.50 x 10-5 /(2 x 8.854 x 10-12 ) 0.04 = 11294 V

Electric field , at the mid point due to line of charge,

E = λ/2πεo R/2= 2.0 x 10-7 / 2 x 3.14 x 8.854 x 10-12 x 0.04 = 8991.4 N/C

Potential difference , dV2 = - Edr = 8991.4 x 0.04 = 359.6 V

Total potential difference between the midpoint and the sheet,

V = dV1 + dV2 = 11294 + 359.6 = 11653.6 V

using law of conservation of energy,

loss electric potential energy = gain of kinetic energy

eV = ½ mv2

charge of electron, e = 1.6 x 10-19 C

mass of electron ,m = 9.11 x 10-31 kg

1.6 x 10-19x 11653.6 = ½ x 9.11 x 10-31 v2

Speed of electron when hit the sheet, v = 4.52 x 107 m/s

4. Given:

C1 = 10 nF = 10 x 10-9 F

C2 =20 nF = 10 x 10-9 F

C3 = 30 nF =10 x 10-9 F

Voltage applied, V = 11 V

(A) The three capacitors are connected in series.

If Cs is the effective capacitance of the combination,

1/Cs = 1/C1 + 1/ C2 + 1/C3 = 1/10 + 1/20 + 1/30 =11/60

Cs = 60/11 nF = 60/11 x 10-9 F

the effective capacitance of the combination, Cs = 60/11 x 10-9 F

charge on each capacitor, q = Cs x V = 60/11 x 10-9 x 11 = 60 x 10-9 C

charge on each capacitor, q = 60 x 10-9 C

energy stored in the system, U = ½ Cs V2 = ½ x 60/11 x 10-9 x 11 x 11 =330 x 10-9 J

energy stored in the system, U =330 x 10-9 J

(B) ( B) The three capacitors are connected in parallel.

If Cp is the effective capacitance of the combination,

Cp = C1 + C2 + C3 = 10 + 20 + 30 = 60 n F = 60 x 10-9 F

the effective capacitance of the combination, Cp = 60x 10-9 F

the charge stored in the network, q = Cp x V = 60 x 10-9 x 11 = 660 x 10-9 C

the charge stored in the network, q = 660 x 10-9 C

the charge on capacitor C1 is q1 = C1 V = 10 x 10-9 x 11 = 110 x 10-9 C

the charge on capacitor C2 is q2 = C2 V = 20 x 10-9 x 11 = 220 x 10-9 C

the charge on capacitor C3 is q3 = C3 V = 30 x 10-9 x 11 = 330 x 10-9 C

energy stored in the system, U = ½ Cp V2 = ½ x 60 x 10-9 x 11 x 11 =3630x 10-9 J

energy stored in the system, U =3630 x 10-9 J