Question

4.35 please

(a) Find the net flux crossing surface r = 2 m and (b) Determine D atr Im and r 5 m 6 m. 4.34 A spherical region of radius a has total charge Q If the charge is uniformly distributed apply Gauss's law to find D both inside and outside the sphere Sections 4.7 and 4.8 Electric Potential and Relationship with iE 4.353) Two point charges Q 2 nC and Q-4nC are located at (1. 0. 3) and (-2.1.5), respectively. Determine the potential at PiI. -2.3) 4.36 A charge of 8 nC is placed at each of the four corners of a square of sides 4 cm long Calculate the electrical potential at the point 3 cm above the center of the square 4.3) (a) total charge Q-60 μC is split into two equal charges located at 180" intervals around a circular loop of radius m Find the potential at the center of the loop (b) If Q is split into three equal charges spaced at 120 intervals around the loop, find the potential at the center ie tfis th limit pr .find the potentalat thee 4.38 Three point charges 0-1 mc.o- 2nC and Q, rnC are, respectively, located at (0,0,4).-2..1)and (3, -4,6) (a) Find the potential y, at P(-1,1,2). tb) Calculate the potential difference V it Qis1.2.3 The potential distribution in free space is given by 4.39 Calculate the electric field strength at (4, π/4. ta: Eat (34.-6) (b) the charge within the cube。<バ1.0くバ1,08 t < 1. 441 the volume charge density inside an atomic nacleus of radius a is p where p, is a constant ta) Calculate the total charge (b) Determine E and V outside the nucleus (c) Determine E and V inside the nocieus (di Prove that E is maximum at0.74Sa omp Design1m lab 1 low ab 7 square numbers.bxt Lab 5.ms14 frequency... b.bmp

Answer 1

4.35

Let Q1=2nC @(1,0,3) and Q2=-4nC @(-2,1,5)

The distance from (1,0,3) to P(1,-2,3) is r1 and (-2,1,5) to P(1,-2,3) is r2.

So r1 is given by r1= (1-1)x + (-2-0)y + (3-3)z = -2y => mod(r1)= 2

r2 is given bt r2=(1-(-2))x + (-2-1)y+ (3-5)z= 3x - 3y -2z => mod(r2)=

Potential at P given by $V= \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}$

Where $k=\frac{1}{4\pi\epsilon_0}=4\pi\times 10^{9}$

Thus $V=4\pi\times 10^{9}(\frac{2n}{2}+\frac{-4n}{\sqrt(22)})= 4\pi\times 10^{9}\times (1-0.85)\times 10^{-9}= 1.84V$