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hodyreva [135]
3 years ago
14

Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.5 µC distributed uniformly on its surface. (a) Find

the electric field 10.0 cm from the center of the charge distribution. (b) Find the electric field 21.5 cm from the center of the charge distribution.
Physics
1 answer:
ra1l [238]3 years ago
8 0

Answer:

10.0     zero, by Gauss' Law the symmetrical distribution will produce no internal electric fields

21.5     E = k Q / R^2     behaves as if all charge were at center

E = 9 E9 * 29.5 E-6 / .215^2 = 5.74E6 N/C

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A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
mafiozo [28]

Answer:

a. f=1.22*10^{-15} N

b. f=53.6*10^{-17} N

Explanation:

The force existing between two charges is given as

f=\frac{kq_{1}q_{2}}{r^{2}}

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

-1.602*10^{-19}

A. we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\  f=1.22*10^{-15} N

b. at  distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\  f=53.6*10^{-17} N

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John, paul, and george are standing in a strawberry field. paul is 14.0 m due west of john. george is 36.0 m from paul, in a dir
sdas [7]

Position of paul with respect to john is given as

14 m due west of john

r_{pj} = r_p - r_j = -14\hat i

position of George with respect to Paul is given as 36 m in direction 37 degree south of east

r_{GP} = r_G - r_p = 36cos37\hat i - 36 sin37\hat j

now we need to find the position of George with respect to John

r_{GJ} = r_G - r_j[\tex]now for the above equation we can add the two equations[tex]r_{Gj} = -14\hat i + 36 cos37\hat i - 36sin37\hat j

r_{Gj} = 14.75\hat i - 21.67\hat j

so the magnitude is given as

r = \sqrt{14.75^2 + 21.67^2} = 26.2 m

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