Calculate the electric field at the following distances from the renter of the s

Question

Calculate the electric field at the following distances from the renter of the sphere, r = R/A, r - R, and r = 3/*/2. Now draw a qualitative graph of the electric field strength versus radial distance for all radii. Draw a qualitative graph of the electric potential versus radial distance for all radii. Beginning from fundamental equations derive the units of capacitance (C) in terms of only the four fundamental units: mass (M), length (L), time (T) and charge (Q). Figure 1 shows the electric potential at the corners of a 1 cm square. What is the electric field-both magnitude and direction-at the center of the square? What is the electric potcutial at the center of the 1 cm square shown in Fig. 2? What is the total electric potential energy stored in this charge configuration? In the circuit shown in Fig. 3. initially the switch is in position A and capacitors C_2 and C_3 are uncharged. After the switch is flipped to position B, what are the charge on and potential difference across each capacitor? The ring of radius R shown in Fig. 4 has a positive charge Q that is uniformly iistributed around its circumference. If a point charge -Q with mass M is placed at the center >f the ring (and given a tiny nudge), what ultimate speed does it obtaiu far out on the axis of he ring? (Use the Work/Energy theorem!)

Explanation / Answer

2.

Capacitance units are Farad ,since

C=Q/V = Coloumb/Volt

since 1 Volt =Joule/Coloumb

C=Coloumb2/Joule

1 Joule =1 Newtion meter

C=Coloumb2/Newton-meter

1 Newton =Kg-meter2/second

C=second2Coloumb2/meter2Kg

---------------------------------------------------------------------------------------------------------------------------------------------

5.

equivalent capacitance of C23 is

1/C23 = 1/3 +1/6

C23 =2 uF

a)

In Position A ,Total Charge on capacitance C1

Q1=C1V =9*4u =36 uC

In Position B ,The new Charge Q1' goes on C1 and the charge Q23 goes on C23.The Voltage across C1 and C23 is same

Q1'+Q23=36 uC-------------1

The Voltage across C1 and C23 is same

Q1'/C1 =Q23/C23

From 1

36-Q23/4 = Q23/2

36-Q23 =2Q23

Q23=12 uC

So Q1' =36-12=24 uC

So Charge on each capacitor is

Q1' =24 uC

Q2 =12 uC

Q3=12 uC

Voltage across each capacitor

V1=Q1/C1 =24/4 =6 Volts

V2=Q2/C2 =12/3 =4 Volts

V3=Q3/C3 =12/6 =2 Volts

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.