Question

Notify Me Bookmark

Calculate the Electric Potential Energy for the next configuration of particles. q1 = 5 C q2 = −4 C q3 = 6 C q4 = −7 C r12 = 0.3 cm r13 = 0.4 cm r14 = 0.8 cm r23 = 0.7 cm r24 = 0.5 cm r34 = 1.2 cm

Calculate the Electric Potential Energy for the next configuration of particles. q1 = 5 C q2 = −4 C q3 = 6 C q4 = −7 C r12 = 0.3 cm r13 = 0.4 cm r14 = 0.8 cm r23 = 0.7 cm r24 = 0.5 cm r34 = 1.2 cm

Image text
  1. Calculate the Electric Potential Energy for the next configuration of particles.
q 1 = 5 C q 2 = − 4 C q 3 = 6 C q 4 = − 7 C r 12 = 0.3 c m r 13 = 0.4 c m r 14 = 0.8 c m r 23 = 0.7 c m r 24 = 0.5 c m r 34 = 1.2 c m

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

Potential in a charge distribution 3 points We have a sphere with diameter d = 7 m and charge density profile ρc(r) = Ar2ω where A = 6×10−8. The change in potential energy of a charge q = 5 mC moved from a radius of r = 2.3 m inside the sphere to the very edge of the sphere was found to be −25 J. Find the value of the constant ω. (Next three questions) We have a spherical ball in a vacuum with three layers: the center and outer layers have constant charge density ρ = 0.7 μC/m3 and the middle layer is a perfect conductor. We'll define zero potential, V = 0, at an infinite distance away from the sphere. The radii are a = 0.3 m, b = 0.7 m, and c = 1 m. Figure 3: Figure for Problems 9-11 Potential at purple dashed line 1 point (a) What is the potential at r = 0.4 m away from the center of the sphere (purple dashed line)? (number 10) Potential at yellow dashed line 1 point What is the potential at r = 0.85 m away from the center of the sphere (yellow dashed line)? (number 11) Potential at green dashed line 1 point What is the potential at r = 1.2 m away from the center of the sphere (green dashed line)?

Solution
0 0

For the following problem, neglect body effect, Cgd and ro, but consider Cgs and use long-channel equations in transistors. Assume L = 0.7 μm for both transistors. The other circuit component values are the following: Rb = 4 kΩ, Rd = 50 Ω, Lg = 10 nH, Ls = 0.5 nH, RS = 50 Ω. The frequency of operation is 2.4 GHz. Ci may be considered as a short circuit at the frequency of operation. Rb is high enough that can be considered as an open circuit for small signal. (a) Symbolically, calculate the input impedance Zi. (b) Calculate the required overdrive voltage in M2 to make the real part of Zi equal to 50 Ω. (c) Calculate W2 to make Zi = 50 Ω. (d) Calculate ID and W1 to make Iref = ID/10. (e) Calculate the DC component at v0. (f) Numerically calculate the small-signal gain, av = vo/vi. (g) Calculate vo if vs = 1 mV peak. (h) Symbolically, calculate the noise figure of this amplifier. (i) Calculate F in dB. CMOS process data: VDD = 1.8 V. For NMOS, Vthn = 0.4V, μn = 400 cm2/Vs, Cox = 9 fF/μm2 and γn = 1. The parameters for PMOS transistors are k′ = μpCox = 90 μA/V2, Vthp = −0.4V and γp = 0.8.
Solution
0 0

Two small objects each of mass m = 0.6 kg are connected by a lightweight rod of length d = 0.7 m (see the figure). At a particular instant they have velocities whose magnitudes are v1 = 37 m/s and v2 = 57 m/s and are subjected to external forces whose magnitudes are F1 = 73 N and F2 = 32 N. The distance h = 0.2 m, and the distance w = 0.8 m. The system is moving in outer space. (a) What is the total (linear) momentum P→total of this system? P→total = kg⋅m/s (b) What is the velocity v→cm of the center of mass? v→cm = m/s (c) What is the total angular momentum L→A of the system relative to point A? L→A = kg⋅m2 /s (d) What is the rotational angular momentum L→rot of the system? L→rot = kg⋅m2 /s (e) What is the translational angular momentum L→trans of the system relative to point A? L→trans = kg⋅m2 /s (f) After a short time interval Δt = 0.23 s, what is the total (linear) momentum P→ total of the system? P→total = kg⋅m/s
Solution
0 0