The voltage across a capacitor discharging into a fixed resistance decays exponentially. The time constant is RC, where C is the capacitance, and R is the resistance between the terminals of the resistor.
$$
V(t)=V(0)e^{\frac{-t}{RC}} \\
$$
You're right that the capacitor starts at 20V, because that's the voltage across the 200 ohm resistor after the capacitor is charged. So you know V(0), and you know C. All you're missing is R. To analyze that properly, think of the switch as a resistor of 0 ohms when closed, and don't worry about the current source. The parallel combination of a 200 ohm resistor and a 0 ohm resistor is 0 ohms. The series combination of a 50 ohm resistor and a 0 ohm resistor is 50 ohms. So the 50 ohm resistor is the only one that matters when determining your discharge constant.
The current through the 200 ohm resistor depends on the voltage across the 200 ohm resistor. The voltage across the 200 ohm resistor is the same as the voltage across the 0 ohm resistor (the closed switch). V=IR, so what's the voltage across that pair of resistors when the switch is closed?
And yes, the voltage across an ideal resistor can change instantaneously. Keep in mind though, there's no such thing as an ideal resistor in the physical world. Everything has capacitance to everything else.
Why is the resistor voltage initially equal to supply voltage? Is it because there is no voltage going across the capacitor yet? Therefore, as there is no voltage drop across the capacitor, all the voltage from the battery is across the resistor?
Sum of voltages on the passive elements must add up to the supply voltage.
$$
V_{supply}(t) = V_{switch}(t) + V_{resistor}(t) + V_{capacitor}(t)
$$
Because of the fact that \$V_{switch}(t) = 0 \$ and \$V_{capacitor}(0) = 0 \$, \$V_{resistor}(0)\$ must be equal to \$V_{supply}(0)\$.
2.What exactly does "the voltage developed as the capacitor charges" refer to?
When you apply a voltage difference between capacitor plates, one plate has more positive potential with respect to the other one. This initiates an electric field field between the plates, which is a vector field, whose direction is from the positive plate the negative one.
There is an insulating material (dielectric material) between these capacitor plates. This dielectric material has no free electrons, so no charge flows through it. But another phenomenon occurs. The negatively charged electrons of the dielectric material tend to the positive plate, while the nucleus of the atoms/molecules shift to the negative plate. This causes a difference in the locations of "center of charge" of electrons and molecules in the dielectric field. This difference create tiny displacement dipols (electric field vectors) inside the dielectric material. This field makes the free electrons in the positive plate go away, while it collects more free electrons to the negative plate. This is how charge is collected in the capacitor plates.
3.Am i correct in assuming that the resistor voltage drops because the capacitor's voltage is increasing? (kirchoff's law where volt rise = volt drop).
As the capacitor voltage increases, the voltage across the resistor will decrease accordingly because of the Kirchoff's Law, which I formulated above. So, yes, you were correct.
1.If the capacitor's voltage is dropping(due to it being discharged), shouldn't the resistor's voltage be increasing due to kirchoff's law? Also,this should therefore INCREASE the current instead of decreasing it, which would then cause the capacitor to discharge even faster?
You are missing the fact that, the source voltage is zero (i.e.; the voltage source is missing) in the discharge circuit. Substitude \$V_{supply}(t)=0\$ in the formula above. The capacitor voltage will be equal to the resistor voltage in reverse polarities during the discharge. Together, they will tend to zero.
Best Answer
You can look at it this way (since you said you want an answer based on math):
The equation for the current through a capacitor is the following:
\$ I_C = C \dfrac{dV}{dt} \$
So at \$ t=0 \$, when the switch is closed, there's a voltage across the capacitor and \$ \frac{dV}{dt} \$ is really large (theoretically infinite) because there is a change in the voltage in technically 'no time,' therefore the slope \$ (\frac{dV}{dt}) \$ approaches infinity at \$ t=0 \$. Since \$ \frac{dV}{dt} \$ is in theory infinity so is the current through the capacitor (they are directly proportional). That explains why it looks as a short at time zero.
As you probably know differentiation does not exist at sharp edges and this is exactly what is happening from a mathematical point of view. Hope it helps!