Part a)
As we know that energy stored inside the capacitor is given as
[tex]U = \frac{1}{2}CV^2[/tex]
for a given capacitor we know
[tex]Q = CV[/tex]
Now we can use it in above equation to find the energy
[tex]U = \frac{1}{2}QV[/tex]
[tex]U = \frac{1}{2}(3.4\times 10^{-6})(24)[/tex]
[tex]U = 40.8\times 10^{-6} J[/tex]
PART b)
If two negative charges are hold near to each other and then released
Then due to mutual repulsion they start moving away from each other
Due to mutual repulsion as the two charges moving away the electrostatic potential energy of two charges will convert into kinetic energy of the two charges.
So here as they move apart kinetic energy will increase while potential energy will decrease
Part c)
As we know that capacitance is given as
[tex]C = \frac{Q}{V}[/tex]
here we know that
[tex]Q = 3\times 10^{-10}C[/tex]
[tex]V = 35 volts[/tex]
[tex]C = \frac{3\times 10^{-10}}{35} [/tex]
[tex]C = 8.6 \times 10^{-12} F[/tex]
a) E=1/2*QV=1/2*3.4*10^(-6)*24=4.08*10^(-5) J
b) They're negative, means that they'll repel. They have electrostatic energy. Then when released this energy converts into mechanical kinetic energy of motion.
c) C=Q/V=8.57*10^(-12) F, or 8.57 pF.
