A total charge of -6.03 nC is uniformly distributed along an insulating curve of

Question

A total charge of -6.03 nC is uniformly distributed along an insulating curve of per length of 1206 nC/m. A Gaussian surface encloses -4.43 nC of that total charge. What is the total length of the curve? Use the non-integral form of Gauss's law to find the net electric flux through Problem I's Gaussian surface. A uniform electric field makes an angle of 56.7 with a flat surface. Thus the electric field makes an angle of 90.0 degree - 56.7 degree = 33.3 degree with a normal to the surface. The area of the surface is 5.55 times 10^-5 m^2. The resulting electric flux through the surface is 44.4 N middot m^2/C. Calculate the magnitude of the electric field. A positive charge is distributed uniformly through a volume of 0.34 m^3, giving a per volume of 50 nC/m^3. A Gaussian surface encloses 0.22 m^3 of that volume. How much charge does the Gaussian surface enclose? We now consider a spherical Gaussian surface of area 4.44 times 10^-4 m^2. There is a concentric spherical -5.55 pC charge distribution completely inside this Gaussian surface. Find E at any point on this surface. Use the integral form of Gauss's law that contains phi. Be sure to show all the steps in your solution for E. Since E is ____ by ____, we can take it out of the integral giving ___ contourintegral _____ = Here E is directed radially _____ and d A is directed radially ____, 50 phi = ____ middot and cos phi = ____, which gives ____ contourintegral ____ = Then contourintegral dA = ____. Solving, E = ____ (symbols) = _____ (numbers) =

Explanation / Answer

1. given, charge, q = -6.03 nC = -6.03*10^-9 C
   charge per uniot lerngth, lambda = 12.06 nC/m = 12.06*10^-9 C/m
   charge enclosed by gaussean surface, qin = -4.43 nC = -4.43*10^-9 c
   TOTAL LENGTH OF THE CURVE = l
   THEN, q/l = lambda
   l = q/lambda = 6.03/12.06 = 0.5 m
2. From gauss' law
   net flux through the gausseran surface, phi = qin/epsilon [ where q in is the enclosed charge by the surface and epsilon is the permittivity of free space]
   epsilon = 8.8541*10^-12
   sp phi = -4.43*10^-9/8.8541*10^-12 = -500.333 Vm
3. angle of electric field with flat surface, theta = 56.7 deg
   angle with normal = 90 - theta
   area of surface, A = 5.55*10^-5 m^2
   electric flux, phi = 44.4 Nm^2/C
   let the magnitude of electric field be E
   then
   EA*cos(90 - theta) = phi ( flux)
   phi = EAsin(theta)
   E = phi/Asin(theta) = 44.4/5.55*10^-5 *sin(56.7) = 957158.356 V/m
4. volume, V = 0.34 m^3
   charge per unit volume, rho = 50 nC/m^3 = 50*10^-9 C/m^3
   volume enclosed by gausseaan surface, V' = 0.22 m^3
   charge enclosed = rho*V' = 50*10^-9*0.22 = 11*10^-9 C = 11 nC

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