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.
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?
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
thank bro !!! but for
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 =
???
Sorry:
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)9.2308e-24
11)option3
Thanx
2) 9.22e-24
11)The flux reverses sign and remains the same in magnitude
To find the amount of charge on the outer surface of the conducting shell, we first need to understand the charge distribution within the system.
Let's start with the insulating sphere. It has a total charge of 10 microCoulombs uniformly distributed throughout its volume. This means that the charge is spread out evenly, resulting in a uniform charge density. We can find the charge density by dividing the total charge by the volume of the sphere.
The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius of the sphere. In this case, we don't have the radius of the sphere, so we need to find it first.
The conducting shell is surrounding the insulating sphere. Since it is a conductor, the charge on its outer surface should be uniformly distributed. We are given that the total charge on the conducting shell is -3 microCoulombs. Since the charge is negative, we can infer that the conducting shell has an excess of electrons.
Now, let's find the radius of the insulating sphere. We know that the conducting shell (with -3 microCoulombs of charge) is surrounding the insulating sphere, which has a total charge of 10 microCoulombs. This means that the net charge inside the conducting shell (insulating sphere + conducting shell) is 10 microCoulombs + (-3 microCoulombs) = 7 microCoulombs.
Since the insulating sphere has a uniform charge distribution, the net charge inside the sphere is equal to the total charge of the sphere. Therefore, the net charge inside the conducting shell is equal to the total charge of the sphere.
Now we can set up an equation using the net charge inside the conducting shell:
(4/3)πr^3 * (charge density) = 7 microCoulombs
Dividing both sides of the equation by (4/3)πr^3, we can solve for the charge density:
(charge density) = 7 microCoulombs / [(4/3)πr^3]
Now that we have the charge density, we can find the total charge on the outer surface of the conducting shell. The outer surface of the conducting shell is equal to the surface of the insulating shell, which has a total charge of 20 microCoulombs uniformly distributed throughout its volume.
To find the charge on the outer surface of the conducting shell, we can multiply the charge density by the volume of the insulating shell. The volume of the insulating shell is given by the formula V = (4/3)π[(outer radius)^3 - (inner radius)^3].
However, we are not given the inner and outer radii of the insulating shell, so we will not be able to calculate the exact charge on the outer surface of the conducting shell without additional information.
Moving on to part b, the electric field at point A halfway between the outer surface of the insulating sphere and the inner surface of the conducting shell:
To calculate the electric field at point A, we need to consider the contributions from each charge distribution.
Since the insulating sphere has a uniform charge distribution, the electric field contribution from it can be calculated using the equation for the electric field due to a uniformly charged sphere:
Electric field (sphere) = (k * Q) / (r^2)
Where k is the electrostatic constant (9 x 10^9 N m^2/C^2), Q is the charge on the sphere, and r is the distance from the center of the sphere to the point where the electric field is being calculated (in this case, point A).
The conducting shell, being a conductor, will have a uniform electric field throughout its interior. However, since point A is outside the conducting shell, the electric field due to it is zero.
The insulating shell also has a uniform charge distribution, so we can calculate the electric field using the same equation as for the sphere:
Electric field (shell) = (k * Q) / (r^2)
Where Q is the charge on the shell and r is the distance from the center of the shell to the point where the electric field is being calculated (in this case, point A).
To find the total electric field at point A, we can sum up the contributions from each charge distribution:
Total electric field at A = Electric field (sphere) + Electric field (shell)
However, since we don't have the exact charges or radii for the sphere and shell, we are unable to calculate the electric field magnitude at point A accurately without additional information.