b. Gauss law: the E in any region is due to the enclosed charge, so E at any point is due solely to the charge within that radius. You just figure E=k*chargeenclosed/distance from center.
a. charges attract. look at the out shell. the sum of the charges of the inner sphere, and the next shell is 10-3 micro, or 7 microC. This means on the inner surface of the outer shell, ther will be -7microC on that surface, and then 27 on the outer surface (total 20 microC).
Nicht whar?
Consider an insulating sphere with 10 microCoulombs of charge uniformly distributed through its volume. The sphere is surrounded by a conducting spherical shell that has a total charge of -3 microCoulombs. Outside the conucting shell is an insulating shell with total charge 20 microCoulombs uniformly distributed through its volume. In the figure below, which shows a cross section of the structure, the darker shaded regions are insulators and the lighter shaded region is the condcutor.
a. How much charge is on the outer surface of the conducting shell in Coulombs?
b. What is the magnitude of the electric field at point A halfway between the outer surface of the insulatng sphere and the inner surface of the conducting shell, if that point is 5 centimeters from the center of the sphere? Give the field magnitude in Newtons/Coulomb.
9 answers
hai bob what the charge enclosed on this formula E=k*chargeenclosed/distance from center. ??? i used 10microC but give wrong ans . and did u got ans for a???? i checked 20microC and give wrong ans
a)enclosed is 10e-6 C + (-3e-3)C = 7e-6 C
b)E(0.05) = k*1e-5/(0.05^2)=3.6e-7 C
b)E(0.05) = k*1e-5/(0.05^2)=3.6e-7 C
thank bro !!! but for
b)is E(0.05) = k*1e-5/(0.05^2)=3.6e7 C
b)is E(0.05) = k*1e-5/(0.05^2)=3.6e7 C
An insulating sphere of mass m and positive charge q is attached to a spring with length h and spring constant ks and is at equilibrium as shown below:
An infinitely long wire with positive linear charge density λ is placed a distance l away from the charged mass at equilibrium as shown below (note that the position of the top of the spring is fixed):
The previous length of the spring was h. What is the new length of the spring in terms of h, q, ke (type "ke"), λ (type "lambda"), l, and ks (type "ks") as needed. Indicate multiplication with a "*" sign and division with a "/" sign. HINT: You can do this without considering the mass or gravitational force.
length of the spring =
???
An infinitely long wire with positive linear charge density λ is placed a distance l away from the charged mass at equilibrium as shown below (note that the position of the top of the spring is fixed):
The previous length of the spring was h. What is the new length of the spring in terms of h, q, ke (type "ke"), λ (type "lambda"), l, and ks (type "ks") as needed. Indicate multiplication with a "*" sign and division with a "/" sign. HINT: You can do this without considering the mass or gravitational force.
length of the spring =
???
Sorry:
For b) is 3.6e7 N/C
For b) is 3.6e7 N/C
1- 19.2E-19
2-???
3-1
4-4.8E-5
5-3.84e-23
6-h-(2*ke*lambda*q)/(l*ks)
7-(E*q)/(m*g)
8-D
9-7e-6
10-3.6e7
11-???
12-3rd option
13-54e9
14-408.5E-12
15-4.902E-9
16-2.451E-9
17-6000
18-96E-17
19-1053787047200878.1558726673984632
20-increase
21-stay the same
22-increase
23-increase
please 2 && 11!!!!
2-???
3-1
4-4.8E-5
5-3.84e-23
6-h-(2*ke*lambda*q)/(l*ks)
7-(E*q)/(m*g)
8-D
9-7e-6
10-3.6e7
11-???
12-3rd option
13-54e9
14-408.5E-12
15-4.902E-9
16-2.451E-9
17-6000
18-96E-17
19-1053787047200878.1558726673984632
20-increase
21-stay the same
22-increase
23-increase
please 2 && 11!!!!
2)9.2308e-24
11)option3
11)option3
Thanx
2) 9.22e-24
11)The flux reverses sign and remains the same in magnitude
2) 9.22e-24
11)The flux reverses sign and remains the same in magnitude
1 answer