Use the following constants if necessary. Coulomb constant, k = 8.987 x 10°N · m2/c2. Vacuum permitivity, E0 = 8.854 x 1012 F/m. Magnetic Permeability of vacuum, uo = 12.566370614356 x 10 -7H/m. Magnitude of the Charge of one electron, e = - 1.60217662 x 10 - 19 c. Mass of one electron, me = 9.10938356 x 10-31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, µC means micro coulomb. p = 68 C/m3 is the volume charge density of a solid sphere of radius R = 5m. Assuming p remains constant throught the sphere. Find the enclosed charge if the radius of the Gaussian surface r = 21 m enclosed charge inside the Gaussian surface Give your answer up to at least three significance digits. C

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Use the following constants if necessary. Coulomb constant, k = 8.987 x 10°N m2/c2. Vacuum permitivity,
En = 8.854 x 10 -12 F/m. Magnetic Permeability of vacuum, un = 12.566370614356 x 10 -7H/m. Magnitude of the
Charge of one electron, e = 1.60217662 x 10 - 19 C. Mass of one electron, m. = 9.10938356 x 10 -31 kg. Unless
specified otherwise, each symbol carries their usual meaning. For example, µc means micro coulomb.
p = 68 C/m3 is the volume charge density of a solid sphere of radius R = 5 m. Assuming p remains constant throught the
sphere. Find the enclosed charge if the radius of the Gaussian surface r = 21 m
enclosed charge inside the Gaussian surface
Give your answer up to at least three significance digits.
C
Transcribed Image Text:Use the following constants if necessary. Coulomb constant, k = 8.987 x 10°N m2/c2. Vacuum permitivity, En = 8.854 x 10 -12 F/m. Magnetic Permeability of vacuum, un = 12.566370614356 x 10 -7H/m. Magnitude of the Charge of one electron, e = 1.60217662 x 10 - 19 C. Mass of one electron, m. = 9.10938356 x 10 -31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, µc means micro coulomb. p = 68 C/m3 is the volume charge density of a solid sphere of radius R = 5 m. Assuming p remains constant throught the sphere. Find the enclosed charge if the radius of the Gaussian surface r = 21 m enclosed charge inside the Gaussian surface Give your answer up to at least three significance digits. C
Gauss' law due to a continuous charge distribution can be written as
1
SSE dA
SSS pdv.
E0
The integral on the right-hand-side of this equation is
a point charge.
a line integral.
a surface integral.
a volume integral.
Transcribed Image Text:Gauss' law due to a continuous charge distribution can be written as 1 SSE dA SSS pdv. E0 The integral on the right-hand-side of this equation is a point charge. a line integral. a surface integral. a volume integral.
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