Answer:
The voltage is
Explanation:
From the question we are told that
The time that has passed is 
Here
is know as the time constant
The voltage of the power source is 
Generally the voltage equation for charging a capacitor is mathematically represented as
![V = V_b [1 - e^{- \frac{t}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Cfrac%7B%5Ctau%7D%7B2%7D%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\tau}{2\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Ctau%7D%7B2%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{1}{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%20%7D%5D)
=>
Answer:
Change in electric potential energy is -28.0 J
Explanation:
Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.
OR
Electric potential energy is also equal to the change in the configuration of the charge particles.
Thus,
Change in electric potential energy = - Work Done
According to the problem, Work Done is equal to 28 J. Thus,
Change in electric potential energy = -28 J