Answer:
a) 50μC
b) 37.45 m/s
Explanation:
a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.
Thus, you have:
Hence, each sphere has a charge of 50μC.
b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:
![\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J](https://tex.z-dn.net/?f=%5CDelta%20W%3D%5CDelta%20K%5C%5C%5C%5C%5Cint_%7B0.4%7D%5E%5Cinfty%20Fdr%3D%5Cfrac%7B1%7D%7B2%7Dm%5Bv%5E2-v_o%5E2%5D%5C%5C%5C%5CF%3Dk%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5C%5C%5C%5Cv_o%3D0m%2Fs%5C%5C%5C%5Cm%3D0.08kg%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3DkQ%5E2%5B-%5Cfrac%7B1%7D%7Br%7D%5D_%7B0.4%7D%5E%7B%5Cinfty%7D%3D%5Cfrac%7BkQ%5E2%7D%7B0.4m%7D%3D%5Cfrac%7B%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%2850%2A10%5E%7B-6%7DC%29%5E2%7D%7B0.4m%7D%5C%5C%5C%5CkQ%5E2%5Cint_%7B0.4%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bdr%7D%7Br%5E2%7D%3D56.125J)
where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2
Next, you equal the total work to the change in K:
hence, the speed of the spheres is 37.45 m/s
Hello yes whats the problem
Answer:
Explanation:
Given that,
The position of a particle is given by :
Let us assume we need to find its velocity.
We know that,
So, the velocity of the particle is 
.
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
Explanation:
If box weight 25N on ground
MA=F
M(10)=25
M=2.5Kg
