Question:
A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =  is located in the center of a spherical cavity of radius ,
 is located in the center of a spherical cavity of radius ,  m inside an insulating spherical charged solid.
  m inside an insulating spherical charged solid.  
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V  =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%2A%5Ctimes%2010%5E%7B-4%20%7D%5B8.57375%20-%202.81011%20%5D%5Ctimes%207.35%5Ctimes%2010%5E%7B-4%7D) 
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at  m due to shell
m due to shell
E1 =  

Electric field at   due to 'q' at center
 due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
= E2- E1
![=[  2.134  - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

 
        
             
        
        
        
Answer:
 a = 1 m/s²  and 
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
       Ty - W = 0         
       Ty = w
X axis
      Tx = m a
With trigonometry we find the components of tension
     Sin θ = Ty / T
     Ty = T sin θ
     Cos θ = Tx / T
     Tx = T cos θ
We calculate the acceleration with kinematics
    Vf = Vo + a t
    a = (Vf -Vo) / t
    a = (20 -10) / 10
    a = 1 m/s²
We substitute in Newton's equations
      
   T Sin θ = mg
   T cos θ = ma
We divide the two equations
   Tan θ = g / a
   θ = tan⁻¹ (g / a)
   θ = tan⁻¹ (9.8 / 1)
   θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
 
        
             
        
        
        
 Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.
 
        
             
        
        
        
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J